Explanation:
Okay, so I need to find the value of k such that (3^k)^6 equals 3^6. Let me think about how to approach this.

First, I remember that when you have an exponent raised to another exponent, like (a^b)^c, you can multiply the exponents. So, (a^b)^c = a^(b*c). Is that right? Let me check with a simple example. If I take (2^3)^2, that should be 2^(3*2) = 2^6, which is 64. And if I calculate it step by step, (2^3) is 8, and 8 squared is 64. Yep, that works. So the rule seems correct.

Applying that to the problem here: (3^k)^6 would be 3^(k*6). So the equation becomes 3^(6k) = 3^6.

Now, since the bases are the same (both are base 3), the exponents must be equal for the equation to hold true. That is, if a^m = a^n, then m = n, provided that a is not 0 or 1. Since 3 is neither 0 nor 1, this rule applies here.

Therefore, setting the exponents equal: 6k = 6.

To solve for k, I can divide both sides of the equation by 6.

So, 6k / 6 = 6 / 6, which simplifies to k = 1.

Wait, let me verify that. If k is 1, then (3^1)^6 = 3^6. Which is 3^6 = 3^6. That's definitely true.

But let me think again if there could be other solutions. For example, sometimes with exponents, especially if dealing with negative numbers or fractions, there might be multiple solutions. But in this case, since the base is 3, which is positive and not equal to 1, the function 3^x is one-to-one. That means each output corresponds to exactly one input. Therefore, there's only one solution for k, which is 1.

Alternatively, if I didn't remember the exponent rules, I could try plugging in numbers. Let's see. If k is 1, then (3^1)^6 = 3^6. If k is 0, then (3^0)^6 = 1^6 = 1, which is not equal to 3^6. If k is 2, then (3^2)^6 = 9^6, which is way bigger than 3^6. So k=1 is the only one that works.

Another way to think about it: 3^6 is 729. Let me compute (3^k)^6 and see when it equals 729.

If k=1: 3^1 = 3, then 3^6 = 729. Perfect.

If k=0: 3^0 = 1, 1^6 = 1 ≠ 729.

k=2: 3^2 = 9, 9^6 = 531441, which is way larger.

So yeah, k=1 is the only solution.

Is there a possibility that k could be a fraction? For example, if k was 1/6, then (3^(1/6))^6 = 3^( (1/6)*6 ) = 3^1 = 3, which is not equal to 3^6. So that's not it.

What if k was something else? Let me think. Suppose k is a negative number. Let's say k = -1. Then (3^-1)^6 = (1/3)^6 = 1/(3^6) which is 1/729, not equal to 729. So no.

Therefore, k must be 1.

Alternatively, maybe using logarithms? Let me try that approach. Taking the logarithm base 3 of both sides.

Starting with (3^k)^6 = 3^6.

Take log base 3: log_3[(3^k)^6] = log_3(3^6).

Left side: log_3(3^(6k)) = 6k.

Right side: log_3(3^6) = 6.

Therefore, 6k = 6 => k=1.

Same result.

So regardless of the method, it seems that k=1 is the solution.

I think that's thorough enough. I considered different approaches, checked for possible alternative solutions, and verified with examples. All lead to k=1.

**Final Answer**
The value of $ k $ is \boxed{1}.

To solve the equation:

$$
(3^k)^6 = 3^6
$$

we begin by applying the **power of a power rule** in exponents, which states:

$$
(a^b)^c = a^{b \cdot c}
$$

Applying this to the left-hand side:

$$
(3^k)^6 = 3^{k \cdot 6} = 3^{6k}
$$

So the equation becomes:

$$
3^{6k} = 3^6
$$

---

Since the bases are the same (both are 3), and the function $ 3^x $ is **one-to-one** (i.e., it is strictly increasing and injective), the exponents must be equal for the equality to hold. Therefore, we equate the exponents:

$$
6k = 6
$$

Now, solve for $ k $ by dividing both sides by 6:

$$
k = \frac{6}{6} = 1
$$

---

To confirm, substitute $ k = 1 $ back into the original expression:

$$
(3^1)^6 = 3^6
$$

Which is clearly true, as both sides are equal to $ 3^6 $.

---

**Final Answer:**

$$
\boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of k such that (3^k)^6 equals 3^6. Let me think about how to approach this.

First, I remember that when you have an exponent raised to another exponent, like (a^b)^c, you can multiply the exponents. So, (a^b)^c = a^(b*c). Is that right? Let me check with a simple example. If I take (2^3)^2, that should be 2^(3*2) = 2^6, which is 64. And if I calculate it step by step, (2^3) is 8, and 8 squared is 64. Yep, that works. So the rule seems correct.

Applying that to the problem here: (3^k)^6 would be 3^(k*6). So the equation becomes 3^(6k) = 3^6.

Now, since the bases are the same (both are base 3), the exponents must be equal for the equation to hold true. That is, if a^m = a^n, then m = n, provided that a is not 0 or 1. Since 3 is neither 0 nor 1, this rule applies here.

Therefore, setting the exponents equal: 6k = 6.

To solve for k, I can divide both sides of the equation by 6.

So, 6k / 6 = 6 / 6, which simplifies to k = 1.

Wait, let me verify that. If k is 1, then (3^1)^6 = 3^6. Which is 3^6 = 3^6. That's definitely true.

But let me think again if there could be other solutions. For example, sometimes with exponents, especially if dealing with negative numbers or fractions, there might be multiple solutions. But in this case, since the base is 3, which is positive and not equal to 1, the function 3^x is one-to-one. That means each output corresponds to exactly one input. Therefore, there's only one solution for k, which is 1.

Alternatively, if I didn't remember the exponent rules, I could try plugging in numbers. Let's see. If k is 1, then (3^1)^6 = 3^6. If k is 0, then (3^0)^6 = 1^6 = 1, which is not equal to 3^6. If k is 2, then (3^2)^6 = 9^6, which is way bigger than 3^6. So k=1 is the only one that works.

Another way to think about it: 3^6 is 729. Let me compute (3^k)^6 and see when it equals 729.

If k=1: 3^1 = 3, then 3^6 = 729. Perfect.

If k=0: 3^0 = 1, 1^6 = 1 ≠ 729.

k=2: 3^2 = 9, 9^6 = 531441, which is way larger.

So yeah, k=1 is the only solution.

Is there a possibility that k could be a fraction? For example, if k was 1/6, then (3^(1/6))^6 = 3^( (1/6)*6 ) = 3^1 = 3, which is not equal to 3^6. So that's not it.

What if k was something else? Let me think. Suppose k is a negative number. Let's say k = -1. Then (3^-1)^6 = (1/3)^6 = 1/(3^6) which is 1/729, not equal to 729. So no.

Therefore, k must be 1.

Alternatively, maybe using logarithms? Let me try that approach. Taking the logarithm base 3 of both sides.

Starting with (3^k)^6 = 3^6.

Take log base 3: log_3[(3^k)^6] = log_3(3^6).

Left side: log_3(3^(6k)) = 6k.

Right side: log_3(3^6) = 6.

Therefore, 6k = 6 => k=1.

Same result.

So regardless of the method, it seems that k=1 is the solution.

I think that's thorough enough. I considered different approaches, checked for possible alternative solutions, and verified with examples. All lead to k=1.

**Final Answer**
The value of $ k $ is \boxed{1}.

To solve the equation:

$$
(3^k)^6 = 3^6
$$

we begin by applying the **power of a power rule** in exponents, which states:

$$
(a^b)^c = a^{b \cdot c}
$$

Applying this to the left-hand side:

$$
(3^k)^6 = 3^{k \cdot 6} = 3^{6k}
$$

So the equation becomes:

$$
3^{6k} = 3^6
$$

---

Since the bases are the same (both are 3), and the function $ 3^x $ is **one-to-one** (i.e., it is strictly increasing and injective), the exponents must be equal for the equality to hold. Therefore, we equate the exponents:

$$
6k = 6
$$

Now, solve for $ k $ by dividing both sides by 6:

$$
k = \frac{6}{6} = 1
$$

---

To confirm, substitute $ k = 1 $ back into the original expression:

$$
(3^1)^6 = 3^6
$$

Which is clearly true, as both sides are equal to $ 3^6 $.

---

**Final Answer:**

$$
\boxed{1}
$$

Solution: Following exponent laws, ${(3^k)}^6=3^{6k}$.  Because $3^{6k}=3^6$, we have $6k=6$, which, dividing by 6, solves to $k=\boxed{1}$.


Explanation:
Okay, so Hillary has eleven coins, and they're all either dimes or nickels. The total value of these coins is 75 cents. I need to figure out how many nickels she has. Let me think about how to approach this.

First, let me recall what dimes and nickels are worth. A dime is 10 cents, and a nickel is 5 cents. So, if I can figure out how many of each coin there are, I can add up their values to get 75 cents. Since there are eleven coins in total, I can set up some equations here.

Let me denote the number of dimes as D and the number of nickels as N. Then, according to the problem, the total number of coins is 11. So, that gives me my first equation:

D + N = 11

And the total value of the coins is 75 cents. Since each dime is 10 cents and each nickel is 5 cents, the second equation would be:

10D + 5N = 75

So now I have a system of two equations with two variables. I can solve this system to find the values of D and N. Let me write them again:

1) D + N = 11

2) 10D + 5N = 75

Hmm, okay. Let me think about how to solve this. Maybe substitution? Or elimination? Let me try substitution. From equation 1, I can express D in terms of N or vice versa. Let me solve equation 1 for D:

D = 11 - N

Then, substitute this expression for D into equation 2. Let me do that:

10*(11 - N) + 5N = 75

Let me compute that:

First, multiply out the 10:

10*11 - 10*N + 5N = 75

Which is:

110 - 10N + 5N = 75

Combine like terms:

110 - 5N = 75

Now, subtract 110 from both sides:

-5N = 75 - 110

Which is:

-5N = -35

Now, divide both sides by -5:

N = (-35)/(-5) = 7

So, N = 7. That means Hillary has 7 nickels. Let me check if this makes sense.

If there are 7 nickels, then the number of dimes would be 11 - 7 = 4. Let me check the total value:

7 nickels * 5 cents = 35 cents

4 dimes * 10 cents = 40 cents

Total is 35 + 40 = 75 cents. Perfect, that matches the total given. So that seems correct.

Wait, but let me just make sure I didn't make any mistakes in my algebra. Let me go through the steps again.

Starting with D + N = 11, so D = 11 - N. Then substituting into the second equation:

10*(11 - N) + 5N = 75

Calculates to 110 - 10N + 5N = 75, which simplifies to 110 - 5N = 75. Then subtract 110: -5N = -35. Divide by -5: N = 7. Yep, that's correct.

Alternatively, maybe I can solve this using elimination. Let me try that method as a check.

Original equations:

1) D + N = 11

2) 10D + 5N = 75

If I multiply equation 1 by 5, I get:

5D + 5N = 55

Then subtract that from equation 2:

(10D + 5N) - (5D + 5N) = 75 - 55

Which gives:

5D = 20

Therefore, D = 4. Then, substituting back into equation 1, N = 11 - 4 = 7. Same result. So that's consistent.

So, both methods give me N = 7. Therefore, Hillary has 7 nickels. Let me just think if there's another way to approach this problem, maybe by trial and error?

Well, since there are 11 coins, and they are either nickels or dimes. Let me think about how many nickels there could be. Let me try different numbers.

Suppose there are 0 nickels: all dimes. Then total value would be 11*10 = 110 cents. That's too much.

If there is 1 nickel, then 10 dimes: total value is 1*5 + 10*10 = 5 + 100 = 105 cents. Still too much.

2 nickels: 9 dimes. 2*5 + 9*10 = 10 + 90 = 100. Still too high.

3 nickels: 8 dimes. 15 + 80 = 95. Still too high.

4 nickels: 7 dimes. 20 + 70 = 90. Still too high.

5 nickels: 6 dimes. 25 + 60 = 85. Still too high.

6 nickels: 5 dimes. 30 + 50 = 80. Still too high.

7 nickels: 4 dimes. 35 + 40 = 75. There we go. That's exactly the total. So that's the answer. So trial and error also leads me to 7 nickels. So that's consistent with the previous methods.

Alternatively, maybe think in terms of the difference between dimes and nickels. Since each dime is worth 5 cents more than a nickel. So if I have D dimes and N nickels, the total value can be thought of as 5*(D + N) + 5D. Wait, let me think.

Wait, each nickel is 5 cents, each dime is 5 cents plus 5 cents. So total value is 5*(number of coins) + 5*(number of dimes). So total value = 5*11 + 5D = 55 + 5D. And this is supposed to equal 75 cents. Therefore:

55 + 5D = 75

Subtract 55: 5D = 20 => D = 4. Therefore, number of dimes is 4, so number of nickels is 11 - 4 = 7. That's another way to think about it. So that's a third method, which also gives the same answer. So that's reassuring.

So all methods lead to 7 nickels. Therefore, I can be confident that the answer is 7.

But just to make sure, let me check once again. If there are 7 nickels, that's 7*5=35 cents. 4 dimes is 4*10=40 cents. 35+40=75. Yep. Perfect. And total coins: 7+4=11. Perfect. So that's correct.

I think that's thorough enough. I considered multiple methods: substitution, elimination, trial and error, and even a different approach thinking about the total value in terms of nickels and the extra from dimes. All lead to the same answer. So I don't see any mistakes here. Therefore, the number of nickels Hillary has is 7.

**Final Answer**
Hillary has \boxed{7} nickels.

To determine how many nickels Hillary has, we are given the following information:

- Hillary has **11 coins** in total.
- All coins are either **dimes (10 cents)** or **nickels (5 cents)**.
- The **total value** of the coins is **75 cents**.

---

### Step 1: Define Variables

Let:
- $ D $ = number of dimes
- $ N $ = number of nickels

From the problem, we can write two equations:

1. **Total number of coins:**
   $$
   D + N = 11
   $$

2. **Total value of coins:**
   $$
   10D + 5N = 75
   $$

---

### Step 2: Solve the System of Equations

We can solve this system using **substitution**.

From equation (1):
$$
D = 11 - N
$$

Substitute into equation (2):
$$
10(11 - N) + 5N = 75
$$

Simplify:
$$
110 - 10N + 5N = 75
$$
$$
110 - 5N = 75
$$

Subtract 110 from both sides:
$$
-5N = -35
$$

Divide both sides by -5:
$$
N = 7
$$

---

### Step 3: Verify the Solution

If $ N = 7 $, then $ D = 11 - 7 = 4 $

- Value of nickels: $ 7 \times 5 = 35 $ cents
- Value of dimes: $ 4 \times 10 = 40 $ cents
- Total value: $ 35 + 40 = 75 $ cents ✅

Also, total number of coins: $ 7 + 4 = 11 $ ✅

---

### Final Answer

$$
\boxed{7}
$$

[Assistant]
Explanation:
Okay, so Hillary has eleven coins, and they're all either dimes or nickels. The total value of these coins is 75 cents. I need to figure out how many nickels she has. Let me think about how to approach this.

First, let me recall what dimes and nickels are worth. A dime is 10 cents, and a nickel is 5 cents. So, if I can figure out how many of each coin there are, I can add up their values to get 75 cents. Since there are eleven coins in total, I can set up some equations here.

Let me denote the number of dimes as D and the number of nickels as N. Then, according to the problem, the total number of coins is 11. So, that gives me my first equation:

D + N = 11

And the total value of the coins is 75 cents. Since each dime is 10 cents and each nickel is 5 cents, the second equation would be:

10D + 5N = 75

So now I have a system of two equations with two variables. I can solve this system to find the values of D and N. Let me write them again:

1) D + N = 11

2) 10D + 5N = 75

Hmm, okay. Let me think about how to solve this. Maybe substitution? Or elimination? Let me try substitution. From equation 1, I can express D in terms of N or vice versa. Let me solve equation 1 for D:

D = 11 - N

Then, substitute this expression for D into equation 2. Let me do that:

10*(11 - N) + 5N = 75

Let me compute that:

First, multiply out the 10:

10*11 - 10*N + 5N = 75

Which is:

110 - 10N + 5N = 75

Combine like terms:

110 - 5N = 75

Now, subtract 110 from both sides:

-5N = 75 - 110

Which is:

-5N = -35

Now, divide both sides by -5:

N = (-35)/(-5) = 7

So, N = 7. That means Hillary has 7 nickels. Let me check if this makes sense.

If there are 7 nickels, then the number of dimes would be 11 - 7 = 4. Let me check the total value:

7 nickels * 5 cents = 35 cents

4 dimes * 10 cents = 40 cents

Total is 35 + 40 = 75 cents. Perfect, that matches the total given. So that seems correct.

Wait, but let me just make sure I didn't make any mistakes in my algebra. Let me go through the steps again.

Starting with D + N = 11, so D = 11 - N. Then substituting into the second equation:

10*(11 - N) + 5N = 75

Calculates to 110 - 10N + 5N = 75, which simplifies to 110 - 5N = 75. Then subtract 110: -5N = -35. Divide by -5: N = 7. Yep, that's correct.

Alternatively, maybe I can solve this using elimination. Let me try that method as a check.

Original equations:

1) D + N = 11

2) 10D + 5N = 75

If I multiply equation 1 by 5, I get:

5D + 5N = 55

Then subtract that from equation 2:

(10D + 5N) - (5D + 5N) = 75 - 55

Which gives:

5D = 20

Therefore, D = 4. Then, substituting back into equation 1, N = 11 - 4 = 7. Same result. So that's consistent.

So, both methods give me N = 7. Therefore, Hillary has 7 nickels. Let me just think if there's another way to approach this problem, maybe by trial and error?

Well, since there are 11 coins, and they are either nickels or dimes. Let me think about how many nickels there could be. Let me try different numbers.

Suppose there are 0 nickels: all dimes. Then total value would be 11*10 = 110 cents. That's too much.

If there is 1 nickel, then 10 dimes: total value is 1*5 + 10*10 = 5 + 100 = 105 cents. Still too much.

2 nickels: 9 dimes. 2*5 + 9*10 = 10 + 90 = 100. Still too high.

3 nickels: 8 dimes. 15 + 80 = 95. Still too high.

4 nickels: 7 dimes. 20 + 70 = 90. Still too high.

5 nickels: 6 dimes. 25 + 60 = 85. Still too high.

6 nickels: 5 dimes. 30 + 50 = 80. Still too high.

7 nickels: 4 dimes. 35 + 40 = 75. There we go. That's exactly the total. So that's the answer. So trial and error also leads me to 7 nickels. So that's consistent with the previous methods.

Alternatively, maybe think in terms of the difference between dimes and nickels. Since each dime is worth 5 cents more than a nickel. So if I have D dimes and N nickels, the total value can be thought of as 5*(D + N) + 5D. Wait, let me think.

Wait, each nickel is 5 cents, each dime is 5 cents plus 5 cents. So total value is 5*(number of coins) + 5*(number of dimes). So total value = 5*11 + 5D = 55 + 5D. And this is supposed to equal 75 cents. Therefore:

55 + 5D = 75

Subtract 55: 5D = 20 => D = 4. Therefore, number of dimes is 4, so number of nickels is 11 - 4 = 7. That's another way to think about it. So that's a third method, which also gives the same answer. So that's reassuring.

So all methods lead to 7 nickels. Therefore, I can be confident that the answer is 7.

But just to make sure, let me check once again. If there are 7 nickels, that's 7*5=35 cents. 4 dimes is 4*10=40 cents. 35+40=75. Yep. Perfect. And total coins: 7+4=11. Perfect. So that's correct.

I think that's thorough enough. I considered multiple methods: substitution, elimination, trial and error, and even a different approach thinking about the total value in terms of nickels and the extra from dimes. All lead to the same answer. So I don't see any mistakes here. Therefore, the number of nickels Hillary has is 7.

**Final Answer**
Hillary has \boxed{7} nickels.

To determine how many nickels Hillary has, we are given the following information:

- Hillary has **11 coins** in total.
- All coins are either **dimes (10 cents)** or **nickels (5 cents)**.
- The **total value** of the coins is **75 cents**.

---

### Step 1: Define Variables

Let:
- $ D $ = number of dimes
- $ N $ = number of nickels

From the problem, we can write two equations:

1. **Total number of coins:**
   $$
   D + N = 11
   $$

2. **Total value of coins:**
   $$
   10D + 5N = 75
   $$

---

### Step 2: Solve the System of Equations

We can solve this system using **substitution**.

From equation (1):
$$
D = 11 - N
$$

Substitute into equation (2):
$$
10(11 - N) + 5N = 75
$$

Simplify:
$$
110 - 10N + 5N = 75
$$
$$
110 - 5N = 75
$$

Subtract 110 from both sides:
$$
-5N = -35
$$

Divide both sides by -5:
$$
N = 7
$$

---

### Step 3: Verify the Solution

If $ N = 7 $, then $ D = 11 - 7 = 4 $

- Value of nickels: $ 7 \times 5 = 35 $ cents
- Value of dimes: $ 4 \times 10 = 40 $ cents
- Total value: $ 35 + 40 = 75 $ cents ✅

Also, total number of coins: $ 7 + 4 = 11 $ ✅

---

### Final Answer

$$
\boxed{7}
$$

Solution:
Let the number of dimes Hillary has be $d$ and the number of nickels she has be $n$. We have the two equations \begin{align*}
d+n&=11\\
10d+5n&=75
\end{align*} (The last equation is in terms of cents.) To make the second equation nicer, we divide both sides by 5 to get $2d+n=15$. From the first given equation, we have $d=11-n$. Substituting this into the simplified second given equation to eliminate $d$, we get $2(11-n)+n=15\Rightarrow n=7$. Thus, Hillary has $\boxed{7}$ nickels.


Explanation:
Okay, so I need to solve this equation: 441 plus 2 times 21 times 19 plus 361 equals x. Let me write that down first to make sure I have it right.

So, the equation is 441 + 2*(21)*(19) + 361 = x. Alright, let me break this down step by step.

First, I should probably handle the multiplication parts before doing the addition. The equation has three terms: 441, 2*(21)*(19), and 361. So, I need to calculate the middle term first.

Let me start with the middle term: 2*(21)*(19). Hmm, multiplication is associative, so I can multiply 2 and 21 first, then multiply the result by 19. Let me do that.

Calculating 2*21: 2*20 is 40, and 2*1 is 2, so 40+2=42. So, 2*21 is 42. Then, I need to multiply that by 19. So, 42*19. Hmm, how do I compute that? Maybe I can break it down further.

I remember that multiplying by 19 is the same as multiplying by 20 and then subtracting 1. So, 42*20 is 840, and then subtract 42, which would be 840 - 42. Let me check that.

Yes, because 19 is 20 - 1, so 42*(20 - 1) = 42*20 - 42*1 = 840 - 42. Let me compute that. 840 minus 40 is 800, and then minus 2 more is 798. So, 42*19 is 798.

So, the middle term is 798.

Now, let me substitute that back into the original equation. So, the equation becomes 441 + 798 + 361 = x.

Now, I need to add these three numbers together. Let me do it step by step.

First, add 441 and 798. Let me compute that.

441 + 798. Let me think. 400 + 700 is 1100. Then, 41 + 98 is... 41 + 98. Let me break that down. 40 + 90 is 130, and 1 + 8 is 9, so 130 + 9 = 139. So, total is 1100 + 139 = 1239. Wait, let me check that again.

Alternatively, maybe adding them straight up. 441 + 798. Let me write it vertically:

  441
+798
------
Let me add the units place: 1 + 8 = 9.

Tens place: 4 + 9 = 13. So, write down 3, carry over 1.

Hundreds place: 4 + 7 = 11, plus the carried over 1 is 12.

So, the total is 1239. Okay, that matches my previous result.

So, 441 + 798 = 1239.

Now, add 361 to that. So, 1239 + 361. Let me compute that.

Again, maybe breaking it down. 1239 + 300 = 1539. Then, 1539 + 61 = 1539 + 60 + 1 = 1599 + 1 = 1600. Wait, is that right? Let me check.

Alternatively, adding vertically:

 1239
+ 361
-------
Let me add units place: 9 + 1 = 10. Write down 0, carry over 1.

Tens place: 3 + 6 = 9, plus the carried over 1 is 10. Write down 0, carry over 1.

Hundreds place: 2 + 3 = 5, plus the carried over 1 is 6.

Thousands place: 1 remains.

So, the total is 1600.

Wait, so 1239 + 361 = 1600? Let me verify with another method.

Alternatively, 1239 + 361. Let me think of 1239 + 361 as 1239 + 300 + 60 + 1.

1239 + 300 = 1539.

1539 + 60 = 1599.

1599 + 1 = 1600.

Yes, that's correct. So, the total is 1600.

Therefore, x equals 1600.

But wait, let me check if I did everything correctly. Let me verify each step again.

Original problem: 441 + 2*(21)*(19) + 361.

First, compute 2*21*19.

2*21 is 42. Then 42*19.

I calculated 42*19 as 798. Let me check that again.

Another way to compute 42*19: 42*(20 - 1) = 42*20 - 42 = 840 - 42 = 798. Correct.

So, that term is 798.

Then, adding 441 + 798 + 361.

First, 441 + 798.

Let me check with another approach. 441 + 798.

I can think of 441 + 798 as (400 + 40 + 1) + (700 + 90 + 8).

Adding 400 + 700 = 1100.

40 + 90 = 130.

1 + 8 = 9.

Total is 1100 + 130 + 9 = 1239. Correct.

Then adding 361.

So, 1239 + 361.

Alternatively, 1239 + 361.

Let me think of 1239 + 361 as (1200 + 39) + (300 + 61) = 1200 + 300 + 39 + 61 = 1500 + 100 = 1600.

Yes, that works.

So, all steps check out. Therefore, x is 1600.

But wait, just to make sure, maybe there's another way to look at the original problem?

Looking back, the original problem is 441 + 2*(21)*(19) + 361.

Wait, 441 is 21 squared, right? 21*21 is 441. And 361 is 19 squared, because 19*19 is 361.

Oh! So, maybe this is a case of a squared term plus twice a product plus another squared term. Which makes me think of the formula for (a + b)^2 = a^2 + 2ab + b^2.

Wait, let me check. If I consider a = 21 and b = 19, then (a + b)^2 = a^2 + 2ab + b^2.

But in our case, the expression is a^2 + 2ab + b^2, which is exactly (a + b)^2.

So, substituting, that would be (21 + 19)^2.

Let me compute that. 21 + 19 is 40. Then, 40 squared is 1600.

Oh! So, that's a much simpler way to compute it.

So, instead of calculating each term separately, recognizing that the entire expression is a perfect square trinomial.

Therefore, x = (21 + 19)^2 = 40^2 = 1600.

So, that's another way to see it.

Therefore, regardless of the method, the answer is 1600.

So, that confirms my previous result.

Therefore, x is 1600.

I think that's solid. Both methods lead to the same answer, so I can be confident that x equals 1600.

**Final Answer**
The value of $ x $ is \boxed{1600}.

To solve the expression:

$$
441 + 2(21)(19) + 361 = x
$$

we can approach this in two ways: either by computing each term step-by-step, or by recognizing a pattern that simplifies the calculation.

---

### **Step-by-Step Computation**

1. **Compute the middle term:**

   $$
   2 \times 21 \times 19
   $$

   First, compute $ 2 \times 21 = 42 $, then multiply by 19:

   $$
   42 \times 19 = 798
   $$

2. **Substitute back into the original expression:**

   $$
   441 + 798 + 361
   $$

3. **Add the numbers:**

   - $ 441 + 798 = 1239 $
   - $ 1239 + 361 = 1600 $

So, the value of $ x $ is:

$$
x = 1600
$$

---

### **Alternative Method: Recognizing a Perfect Square**

Observe that:

- $ 441 = 21^2 $
- $ 361 = 19^2 $
- The middle term is $ 2 \times 21 \times 19 $

This matches the form of the **square of a binomial**:

$$
(a + b)^2 = a^2 + 2ab + b^2
$$

So, we can rewrite the expression as:

$$
(21 + 19)^2 = 40^2 = 1600
$$

---

### **Final Answer**

$$
\boxed{1600}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: 441 plus 2 times 21 times 19 plus 361 equals x. Let me write that down first to make sure I have it right.

So, the equation is 441 + 2*(21)*(19) + 361 = x. Alright, let me break this down step by step.

First, I should probably handle the multiplication parts before doing the addition. The equation has three terms: 441, 2*(21)*(19), and 361. So, I need to calculate the middle term first.

Let me start with the middle term: 2*(21)*(19). Hmm, multiplication is associative, so I can multiply 2 and 21 first, then multiply the result by 19. Let me do that.

Calculating 2*21: 2*20 is 40, and 2*1 is 2, so 40+2=42. So, 2*21 is 42. Then, I need to multiply that by 19. So, 42*19. Hmm, how do I compute that? Maybe I can break it down further.

I remember that multiplying by 19 is the same as multiplying by 20 and then subtracting 1. So, 42*20 is 840, and then subtract 42, which would be 840 - 42. Let me check that.

Yes, because 19 is 20 - 1, so 42*(20 - 1) = 42*20 - 42*1 = 840 - 42. Let me compute that. 840 minus 40 is 800, and then minus 2 more is 798. So, 42*19 is 798.

So, the middle term is 798.

Now, let me substitute that back into the original equation. So, the equation becomes 441 + 798 + 361 = x.

Now, I need to add these three numbers together. Let me do it step by step.

First, add 441 and 798. Let me compute that.

441 + 798. Let me think. 400 + 700 is 1100. Then, 41 + 98 is... 41 + 98. Let me break that down. 40 + 90 is 130, and 1 + 8 is 9, so 130 + 9 = 139. So, total is 1100 + 139 = 1239. Wait, let me check that again.

Alternatively, maybe adding them straight up. 441 + 798. Let me write it vertically:

  441
+798
------
Let me add the units place: 1 + 8 = 9.

Tens place: 4 + 9 = 13. So, write down 3, carry over 1.

Hundreds place: 4 + 7 = 11, plus the carried over 1 is 12.

So, the total is 1239. Okay, that matches my previous result.

So, 441 + 798 = 1239.

Now, add 361 to that. So, 1239 + 361. Let me compute that.

Again, maybe breaking it down. 1239 + 300 = 1539. Then, 1539 + 61 = 1539 + 60 + 1 = 1599 + 1 = 1600. Wait, is that right? Let me check.

Alternatively, adding vertically:

 1239
+ 361
-------
Let me add units place: 9 + 1 = 10. Write down 0, carry over 1.

Tens place: 3 + 6 = 9, plus the carried over 1 is 10. Write down 0, carry over 1.

Hundreds place: 2 + 3 = 5, plus the carried over 1 is 6.

Thousands place: 1 remains.

So, the total is 1600.

Wait, so 1239 + 361 = 1600? Let me verify with another method.

Alternatively, 1239 + 361. Let me think of 1239 + 361 as 1239 + 300 + 60 + 1.

1239 + 300 = 1539.

1539 + 60 = 1599.

1599 + 1 = 1600.

Yes, that's correct. So, the total is 1600.

Therefore, x equals 1600.

But wait, let me check if I did everything correctly. Let me verify each step again.

Original problem: 441 + 2*(21)*(19) + 361.

First, compute 2*21*19.

2*21 is 42. Then 42*19.

I calculated 42*19 as 798. Let me check that again.

Another way to compute 42*19: 42*(20 - 1) = 42*20 - 42 = 840 - 42 = 798. Correct.

So, that term is 798.

Then, adding 441 + 798 + 361.

First, 441 + 798.

Let me check with another approach. 441 + 798.

I can think of 441 + 798 as (400 + 40 + 1) + (700 + 90 + 8).

Adding 400 + 700 = 1100.

40 + 90 = 130.

1 + 8 = 9.

Total is 1100 + 130 + 9 = 1239. Correct.

Then adding 361.

So, 1239 + 361.

Alternatively, 1239 + 361.

Let me think of 1239 + 361 as (1200 + 39) + (300 + 61) = 1200 + 300 + 39 + 61 = 1500 + 100 = 1600.

Yes, that works.

So, all steps check out. Therefore, x is 1600.

But wait, just to make sure, maybe there's another way to look at the original problem?

Looking back, the original problem is 441 + 2*(21)*(19) + 361.

Wait, 441 is 21 squared, right? 21*21 is 441. And 361 is 19 squared, because 19*19 is 361.

Oh! So, maybe this is a case of a squared term plus twice a product plus another squared term. Which makes me think of the formula for (a + b)^2 = a^2 + 2ab + b^2.

Wait, let me check. If I consider a = 21 and b = 19, then (a + b)^2 = a^2 + 2ab + b^2.

But in our case, the expression is a^2 + 2ab + b^2, which is exactly (a + b)^2.

So, substituting, that would be (21 + 19)^2.

Let me compute that. 21 + 19 is 40. Then, 40 squared is 1600.

Oh! So, that's a much simpler way to compute it.

So, instead of calculating each term separately, recognizing that the entire expression is a perfect square trinomial.

Therefore, x = (21 + 19)^2 = 40^2 = 1600.

So, that's another way to see it.

Therefore, regardless of the method, the answer is 1600.

So, that confirms my previous result.

Therefore, x is 1600.

I think that's solid. Both methods lead to the same answer, so I can be confident that x equals 1600.

**Final Answer**
The value of $ x $ is \boxed{1600}.

To solve the expression:

$$
441 + 2(21)(19) + 361 = x
$$

we can approach this in two ways: either by computing each term step-by-step, or by recognizing a pattern that simplifies the calculation.

---

### **Step-by-Step Computation**

1. **Compute the middle term:**

   $$
   2 \times 21 \times 19
   $$

   First, compute $ 2 \times 21 = 42 $, then multiply by 19:

   $$
   42 \times 19 = 798
   $$

2. **Substitute back into the original expression:**

   $$
   441 + 798 + 361
   $$

3. **Add the numbers:**

   - $ 441 + 798 = 1239 $
   - $ 1239 + 361 = 1600 $

So, the value of $ x $ is:

$$
x = 1600
$$

---

### **Alternative Method: Recognizing a Perfect Square**

Observe that:

- $ 441 = 21^2 $
- $ 361 = 19^2 $
- The middle term is $ 2 \times 21 \times 19 $

This matches the form of the **square of a binomial**:

$$
(a + b)^2 = a^2 + 2ab + b^2
$$

So, we can rewrite the expression as:

$$
(21 + 19)^2 = 40^2 = 1600
$$

---

### **Final Answer**

$$
\boxed{1600}
$$

Solution: We note that $361=19^2$ and $441=21^2$, so $x=21^2+2(21)(19)+19^2$. This is just the binomial expansion of $(21+19)^2=40^2=\boxed{1600}$.


Explanation:
Okay, so I need to find the value of x + y + z given that xy = 24, xz = 48, and yz = 72. All variables are positive. Hmm, let me think about how to approach this.

First, I remember that when you have multiple equations with products of variables, sometimes multiplying all the equations together can help. Let me try that. If I multiply xy, xz, and yz together, that would be (xy)(xz)(yz). Let me compute that:

(xy)(xz)(yz) = x^2 y^2 z^2.

On the other hand, multiplying the right-hand sides of the equations: 24 * 48 * 72.

So, x^2 y^2 z^2 = 24 * 48 * 72.

Therefore, if I take the square root of both sides, I get xyz = sqrt(24 * 48 * 72).

But maybe I can compute that. Let me calculate the right-hand side first. Let me compute 24 * 48 first. 24*48... Well, 24*40 is 960, and 24*8 is 192, so total is 960 + 192 = 1152. Then multiply that by 72. Hmm, 1152 * 72. That seems big. Maybe there's a smarter way to factor these numbers?

Alternatively, maybe I can factor each of the numbers into primes. Let me try that.

24 factors into 2^3 * 3^1.

48 is 2^4 * 3^1.

72 is 2^3 * 3^2.

So multiplying them all together:

24 * 48 * 72 = (2^3 * 3^1) * (2^4 * 3^1) * (2^3 * 3^2)

Adding exponents for like bases:

For 2: 3 + 4 + 3 = 10

For 3: 1 + 1 + 2 = 4

So total is 2^10 * 3^4.

Therefore, sqrt(2^10 * 3^4) = 2^(5) * 3^(2) = 32 * 9 = 288.

So xyz = 288.

Okay, so now I know that xyz = 288.

But how does that help me find x, y, z individually? Well, if I can find each variable by dividing by the product of the other two.

For example, since xy = 24, then z = (xyz)/(xy) = 288 / 24 = 12.

Similarly, x = (xyz)/(yz) = 288 / 72 = 4.

And y = (xyz)/(xz) = 288 / 48 = 6.

Wait, so x is 4, y is 6, z is 12? Let me check if these values satisfy the original equations.

Check xy: 4*6 = 24. Correct.

Check xz: 4*12 = 48. Correct.

Check yz: 6*12 = 72. Correct.

So that works. Then x + y + z = 4 + 6 + 12 = 22.

Wait, so is the answer 22? Let me just make sure I didn't make any calculation errors.

Let me recheck the multiplication steps.

Original equations: xy=24, xz=48, yz=72.

Multiplying all three: (xy)(xz)(yz) = 24*48*72.

Which is x^2 y^2 z^2 = 24*48*72.

Taking square root: xyz = sqrt(24*48*72).

Then factoring each number:

24 = 2^3 * 3

48 = 2^4 * 3

72 = 2^3 * 3^2

Multiply them: 2^(3+4+3) * 3^(1+1+2) = 2^10 * 3^4.

Square root is 2^5 * 3^2 = 32 * 9 = 288.

Yes, that's correct.

Then xyz = 288.

So z = 288 / (xy) = 288 / 24 = 12.

Similarly, x = 288 / (yz) = 288 / 72 = 4.

y = 288 / (xz) = 288 / 48 = 6.

Adding them up: 4 + 6 + 12 = 22.

Seems correct.

Alternatively, maybe I can solve the equations step by step without multiplying all three. Let me try another approach to verify.

From the first equation, xy = 24. Let me solve for y: y = 24/x.

From the second equation, xz = 48. Solve for z: z = 48/x.

From the third equation, yz = 72. Substitute y and z from above: (24/x)*(48/x) = 72.

Compute that: (24*48)/x² = 72.

So 24*48 = 72x².

Calculate 24*48: 24*48. Let me compute 24*48 again. 24*40=960, 24*8=192, total 960+192=1152.

So 1152 = 72x².

Divide both sides by 72: 1152 / 72 = x².

Calculate 1152 / 72. Let me divide numerator and denominator by 8: 144 / 9 = 16.

Wait, 72*16 = 1152? Let me check: 70*16=1120, 2*16=32, so 1120+32=1152. Yes.

So x² = 16, so x = 4 (since x is positive).

Then, from earlier, y = 24/x = 24/4 = 6.

z = 48/x = 48/4 = 12.

Same result. So x + y + z = 4 + 6 + 12 = 22.

So that's consistent.

Alternatively, maybe I can use ratios. Let me see.

From xy = 24 and xz = 48, if I divide the second equation by the first, I get (xz)/(xy) = 48/24 => z/y = 2 => z = 2y.

Similarly, from yz = 72, if z = 2y, then y*(2y) = 72 => 2y² = 72 => y² = 36 => y = 6 (positive).

Then z = 2y = 12. Then from xy = 24, x = 24 / y = 24 / 6 = 4.

Same result again.

So x = 4, y = 6, z = 12. Sum is 22.

Therefore, regardless of the method, I get the same answer.

I think that's solid. So the answer is 22.

**Final Answer**
\boxed{22}

To solve the problem, we are given the following equations involving three positive variables $ x $, $ y $, and $ z $:

$$
xy = 24, \quad xz = 48, \quad yz = 72
$$

Our goal is to find the value of $ x + y + z $.

---

### Step 1: Multiply All Three Equations

Multiplying the left-hand sides and the right-hand sides of the equations:

$$
(xy)(xz)(yz) = 24 \cdot 48 \cdot 72
$$

The left-hand side becomes:

$$
x^2 y^2 z^2
$$

So:

$$
x^2 y^2 z^2 = 24 \cdot 48 \cdot 72
$$

Taking the square root of both sides:

$$
xyz = \sqrt{24 \cdot 48 \cdot 72}
$$

---

### Step 2: Simplify the Product

We simplify the product on the right-hand side by factoring each number into its prime components:

- $ 24 = 2^3 \cdot 3 $
- $ 48 = 2^4 \cdot 3 $
- $ 72 = 2^3 \cdot 3^2 $

Multiplying them:

$$
24 \cdot 48 \cdot 72 = (2^3 \cdot 3) \cdot (2^4 \cdot 3) \cdot (2^3 \cdot 3^2) = 2^{3+4+3} \cdot 3^{1+1+2} = 2^{10} \cdot 3^4
$$

Taking the square root:

$$
xyz = \sqrt{2^{10} \cdot 3^4} = 2^5 \cdot 3^2 = 32 \cdot 9 = 288
$$

So:

$$
xyz = 288
$$

---

### Step 3: Solve for Each Variable

We can now find each variable by dividing the product $ xyz $ by the product of the other two variables:

- $ x = \frac{xyz}{yz} = \frac{288}{72} = 4 $
- $ y = \frac{xyz}{xz} = \frac{288}{48} = 6 $
- $ z = \frac{xyz}{xy} = \frac{288}{24} = 12 $

---

### Step 4: Compute the Sum

$$
x + y + z = 4 + 6 + 12 = \boxed{22}
$$

---

### Final Answer

$$
\boxed{22}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of x + y + z given that xy = 24, xz = 48, and yz = 72. All variables are positive. Hmm, let me think about how to approach this.

First, I remember that when you have multiple equations with products of variables, sometimes multiplying all the equations together can help. Let me try that. If I multiply xy, xz, and yz together, that would be (xy)(xz)(yz). Let me compute that:

(xy)(xz)(yz) = x^2 y^2 z^2.

On the other hand, multiplying the right-hand sides of the equations: 24 * 48 * 72.

So, x^2 y^2 z^2 = 24 * 48 * 72.

Therefore, if I take the square root of both sides, I get xyz = sqrt(24 * 48 * 72).

But maybe I can compute that. Let me calculate the right-hand side first. Let me compute 24 * 48 first. 24*48... Well, 24*40 is 960, and 24*8 is 192, so total is 960 + 192 = 1152. Then multiply that by 72. Hmm, 1152 * 72. That seems big. Maybe there's a smarter way to factor these numbers?

Alternatively, maybe I can factor each of the numbers into primes. Let me try that.

24 factors into 2^3 * 3^1.

48 is 2^4 * 3^1.

72 is 2^3 * 3^2.

So multiplying them all together:

24 * 48 * 72 = (2^3 * 3^1) * (2^4 * 3^1) * (2^3 * 3^2)

Adding exponents for like bases:

For 2: 3 + 4 + 3 = 10

For 3: 1 + 1 + 2 = 4

So total is 2^10 * 3^4.

Therefore, sqrt(2^10 * 3^4) = 2^(5) * 3^(2) = 32 * 9 = 288.

So xyz = 288.

Okay, so now I know that xyz = 288.

But how does that help me find x, y, z individually? Well, if I can find each variable by dividing by the product of the other two.

For example, since xy = 24, then z = (xyz)/(xy) = 288 / 24 = 12.

Similarly, x = (xyz)/(yz) = 288 / 72 = 4.

And y = (xyz)/(xz) = 288 / 48 = 6.

Wait, so x is 4, y is 6, z is 12? Let me check if these values satisfy the original equations.

Check xy: 4*6 = 24. Correct.

Check xz: 4*12 = 48. Correct.

Check yz: 6*12 = 72. Correct.

So that works. Then x + y + z = 4 + 6 + 12 = 22.

Wait, so is the answer 22? Let me just make sure I didn't make any calculation errors.

Let me recheck the multiplication steps.

Original equations: xy=24, xz=48, yz=72.

Multiplying all three: (xy)(xz)(yz) = 24*48*72.

Which is x^2 y^2 z^2 = 24*48*72.

Taking square root: xyz = sqrt(24*48*72).

Then factoring each number:

24 = 2^3 * 3

48 = 2^4 * 3

72 = 2^3 * 3^2

Multiply them: 2^(3+4+3) * 3^(1+1+2) = 2^10 * 3^4.

Square root is 2^5 * 3^2 = 32 * 9 = 288.

Yes, that's correct.

Then xyz = 288.

So z = 288 / (xy) = 288 / 24 = 12.

Similarly, x = 288 / (yz) = 288 / 72 = 4.

y = 288 / (xz) = 288 / 48 = 6.

Adding them up: 4 + 6 + 12 = 22.

Seems correct.

Alternatively, maybe I can solve the equations step by step without multiplying all three. Let me try another approach to verify.

From the first equation, xy = 24. Let me solve for y: y = 24/x.

From the second equation, xz = 48. Solve for z: z = 48/x.

From the third equation, yz = 72. Substitute y and z from above: (24/x)*(48/x) = 72.

Compute that: (24*48)/x² = 72.

So 24*48 = 72x².

Calculate 24*48: 24*48. Let me compute 24*48 again. 24*40=960, 24*8=192, total 960+192=1152.

So 1152 = 72x².

Divide both sides by 72: 1152 / 72 = x².

Calculate 1152 / 72. Let me divide numerator and denominator by 8: 144 / 9 = 16.

Wait, 72*16 = 1152? Let me check: 70*16=1120, 2*16=32, so 1120+32=1152. Yes.

So x² = 16, so x = 4 (since x is positive).

Then, from earlier, y = 24/x = 24/4 = 6.

z = 48/x = 48/4 = 12.

Same result. So x + y + z = 4 + 6 + 12 = 22.

So that's consistent.

Alternatively, maybe I can use ratios. Let me see.

From xy = 24 and xz = 48, if I divide the second equation by the first, I get (xz)/(xy) = 48/24 => z/y = 2 => z = 2y.

Similarly, from yz = 72, if z = 2y, then y*(2y) = 72 => 2y² = 72 => y² = 36 => y = 6 (positive).

Then z = 2y = 12. Then from xy = 24, x = 24 / y = 24 / 6 = 4.

Same result again.

So x = 4, y = 6, z = 12. Sum is 22.

Therefore, regardless of the method, I get the same answer.

I think that's solid. So the answer is 22.

**Final Answer**
\boxed{22}

To solve the problem, we are given the following equations involving three positive variables $ x $, $ y $, and $ z $:

$$
xy = 24, \quad xz = 48, \quad yz = 72
$$

Our goal is to find the value of $ x + y + z $.

---

### Step 1: Multiply All Three Equations

Multiplying the left-hand sides and the right-hand sides of the equations:

$$
(xy)(xz)(yz) = 24 \cdot 48 \cdot 72
$$

The left-hand side becomes:

$$
x^2 y^2 z^2
$$

So:

$$
x^2 y^2 z^2 = 24 \cdot 48 \cdot 72
$$

Taking the square root of both sides:

$$
xyz = \sqrt{24 \cdot 48 \cdot 72}
$$

---

### Step 2: Simplify the Product

We simplify the product on the right-hand side by factoring each number into its prime components:

- $ 24 = 2^3 \cdot 3 $
- $ 48 = 2^4 \cdot 3 $
- $ 72 = 2^3 \cdot 3^2 $

Multiplying them:

$$
24 \cdot 48 \cdot 72 = (2^3 \cdot 3) \cdot (2^4 \cdot 3) \cdot (2^3 \cdot 3^2) = 2^{3+4+3} \cdot 3^{1+1+2} = 2^{10} \cdot 3^4
$$

Taking the square root:

$$
xyz = \sqrt{2^{10} \cdot 3^4} = 2^5 \cdot 3^2 = 32 \cdot 9 = 288
$$

So:

$$
xyz = 288
$$

---

### Step 3: Solve for Each Variable

We can now find each variable by dividing the product $ xyz $ by the product of the other two variables:

- $ x = \frac{xyz}{yz} = \frac{288}{72} = 4 $
- $ y = \frac{xyz}{xz} = \frac{288}{48} = 6 $
- $ z = \frac{xyz}{xy} = \frac{288}{24} = 12 $

---

### Step 4: Compute the Sum

$$
x + y + z = 4 + 6 + 12 = \boxed{22}
$$

---

### Final Answer

$$
\boxed{22}
$$

Solution:
Multiplying all three equations, we get $x^2 y^2 z^2 = 82944.$  Since $x,$$y,$ and $z$ are positive, $xyz = \sqrt{82944} = 288.$  Then
\begin{align*}
x &= \frac{xyz}{yz} = \frac{288}{72} = 4, \\
y &= \frac{xyz}{xz} = \frac{288}{48} = 6, \\
z &= \frac{xyz}{xy} = \frac{288}{24} = 12.
\end{align*}Hence, $x + y + z = \boxed{22}.$


Explanation:
Okay, so I need to compute the product of the number 5 plus the square root of 3 and its radical conjugate. Hmm, let me think. First, what exactly is a radical conjugate? I remember that for expressions involving radicals, the radical conjugate is formed by changing the sign between the terms. So if I have something like a + b√c, its radical conjugate would be a - b√c, right? Let me confirm that. Yeah, I think that's correct. For example, the radical conjugate of 2 + 3√5 would be 2 - 3√5. So applying that to the given problem, the number is 5 + √3, so its radical conjugate should be 5 - √3.

Alright, so the problem is asking for the product of these two numbers: (5 + √3) multiplied by (5 - √3). That seems straightforward. Let me write that down:

(5 + √3)(5 - √3)

Now, I need to multiply these two binomials. I remember there's a formula for multiplying conjugates, which is (a + b)(a - b) = a² - b². Is that right? Let me check with numbers. If a is 2 and b is 3, then (2 + 3)(2 - 3) = (5)(-1) = -5, and using the formula, it would be 2² - 3² = 4 - 9 = -5. Yep, that works. So applying that formula here should simplify things.

In this case, a is 5 and b is √3. So substituting into the formula, we get:

(5)² - (√3)²

Calculating each term separately. First, 5 squared is 25. Then, (√3) squared is just 3, because the square and the square root cancel each other out. So substituting those back in:

25 - 3 = 22

Wait, so the product is 22? That seems too simple. Let me verify by expanding the original expression without using the formula to make sure I didn't make a mistake.

Expanding (5 + √3)(5 - √3):

First, multiply 5 by 5, which is 25.

Then, 5 multiplied by -√3 is -5√3.

Next, √3 multiplied by 5 is +5√3.

Finally, √3 multiplied by -√3 is - (√3)², which is -3.

So putting all those terms together:

25 - 5√3 + 5√3 - 3

Now, combining like terms. The -5√3 and +5√3 cancel each other out, right? Because -5√3 + 5√3 is 0. Then we have 25 - 3, which is 22. Yep, same result. So that checks out.

Therefore, the product of 5 + √3 and its radical conjugate 5 - √3 is indeed 22.

But just to make sure I didn't confuse radical conjugate with something else. Let me recall. The radical conjugate is specifically for expressions with a single radical term. If there were multiple radicals, the conjugate would be different, but in this case, since it's just √3, changing the sign between the terms is correct. For example, if the expression was 5 + √3 + √2, the conjugate would be more complicated, but here it's only one radical.

Another way to think about it is that the radical conjugate is used to rationalize denominators. For instance, if you have a fraction with a denominator like 5 + √3, multiplying numerator and denominator by the radical conjugate 5 - √3 would rationalize the denominator. But in this case, we're just multiplying the two numbers, so the same principle applies.

So, in conclusion, using either the difference of squares formula or expanding the terms manually, I arrive at the same answer, 22. Therefore, I can be confident that the product is 22.

**Final Answer**
The product of the number $5+\sqrt{3}$ and its radical conjugate is \boxed{22}.

To compute the product of the number $ 5 + \sqrt{3} $ and its **radical conjugate**, we first identify the radical conjugate of the expression.

---

### Step 1: Identify the Radical Conjugate

The **radical conjugate** of an expression of the form $ a + b\sqrt{c} $ is $ a - b\sqrt{c} $. In this case, the given number is:

$$
5 + \sqrt{3}
$$

So, its radical conjugate is:

$$
5 - \sqrt{3}
$$

---

### Step 2: Multiply the Expression by Its Conjugate

We now compute the product:

$$
(5 + \sqrt{3})(5 - \sqrt{3})
$$

This is a classic case of the **difference of squares** identity:

$$
(a + b)(a - b) = a^2 - b^2
$$

Here, $ a = 5 $ and $ b = \sqrt{3} $, so:

$$
(5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2
$$

---

### Step 3: Simplify the Expression

$$
5^2 = 25 \quad \text{and} \quad (\sqrt{3})^2 = 3
$$

So:

$$
25 - 3 = 22
$$

---

### Final Answer

$$
\boxed{22}
$$

[Assistant]
Explanation:
Okay, so I need to compute the product of the number 5 plus the square root of 3 and its radical conjugate. Hmm, let me think. First, what exactly is a radical conjugate? I remember that for expressions involving radicals, the radical conjugate is formed by changing the sign between the terms. So if I have something like a + b√c, its radical conjugate would be a - b√c, right? Let me confirm that. Yeah, I think that's correct. For example, the radical conjugate of 2 + 3√5 would be 2 - 3√5. So applying that to the given problem, the number is 5 + √3, so its radical conjugate should be 5 - √3.

Alright, so the problem is asking for the product of these two numbers: (5 + √3) multiplied by (5 - √3). That seems straightforward. Let me write that down:

(5 + √3)(5 - √3)

Now, I need to multiply these two binomials. I remember there's a formula for multiplying conjugates, which is (a + b)(a - b) = a² - b². Is that right? Let me check with numbers. If a is 2 and b is 3, then (2 + 3)(2 - 3) = (5)(-1) = -5, and using the formula, it would be 2² - 3² = 4 - 9 = -5. Yep, that works. So applying that formula here should simplify things.

In this case, a is 5 and b is √3. So substituting into the formula, we get:

(5)² - (√3)²

Calculating each term separately. First, 5 squared is 25. Then, (√3) squared is just 3, because the square and the square root cancel each other out. So substituting those back in:

25 - 3 = 22

Wait, so the product is 22? That seems too simple. Let me verify by expanding the original expression without using the formula to make sure I didn't make a mistake.

Expanding (5 + √3)(5 - √3):

First, multiply 5 by 5, which is 25.

Then, 5 multiplied by -√3 is -5√3.

Next, √3 multiplied by 5 is +5√3.

Finally, √3 multiplied by -√3 is - (√3)², which is -3.

So putting all those terms together:

25 - 5√3 + 5√3 - 3

Now, combining like terms. The -5√3 and +5√3 cancel each other out, right? Because -5√3 + 5√3 is 0. Then we have 25 - 3, which is 22. Yep, same result. So that checks out.

Therefore, the product of 5 + √3 and its radical conjugate 5 - √3 is indeed 22.

But just to make sure I didn't confuse radical conjugate with something else. Let me recall. The radical conjugate is specifically for expressions with a single radical term. If there were multiple radicals, the conjugate would be different, but in this case, since it's just √3, changing the sign between the terms is correct. For example, if the expression was 5 + √3 + √2, the conjugate would be more complicated, but here it's only one radical.

Another way to think about it is that the radical conjugate is used to rationalize denominators. For instance, if you have a fraction with a denominator like 5 + √3, multiplying numerator and denominator by the radical conjugate 5 - √3 would rationalize the denominator. But in this case, we're just multiplying the two numbers, so the same principle applies.

So, in conclusion, using either the difference of squares formula or expanding the terms manually, I arrive at the same answer, 22. Therefore, I can be confident that the product is 22.

**Final Answer**
The product of the number $5+\sqrt{3}$ and its radical conjugate is \boxed{22}.

To compute the product of the number $ 5 + \sqrt{3} $ and its **radical conjugate**, we first identify the radical conjugate of the expression.

---

### Step 1: Identify the Radical Conjugate

The **radical conjugate** of an expression of the form $ a + b\sqrt{c} $ is $ a - b\sqrt{c} $. In this case, the given number is:

$$
5 + \sqrt{3}
$$

So, its radical conjugate is:

$$
5 - \sqrt{3}
$$

---

### Step 2: Multiply the Expression by Its Conjugate

We now compute the product:

$$
(5 + \sqrt{3})(5 - \sqrt{3})
$$

This is a classic case of the **difference of squares** identity:

$$
(a + b)(a - b) = a^2 - b^2
$$

Here, $ a = 5 $ and $ b = \sqrt{3} $, so:

$$
(5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2
$$

---

### Step 3: Simplify the Expression

$$
5^2 = 25 \quad \text{and} \quad (\sqrt{3})^2 = 3
$$

So:

$$
25 - 3 = 22
$$

---

### Final Answer

$$
\boxed{22}
$$

Solution: The radical conjugate of this number is $5-\sqrt{3},$ so the product of the two numbers is \[(5+\sqrt3)(5-\sqrt3) = 5^2 - (\sqrt3)^2 = 25 - 3 = \boxed{22}.\]


Explanation:
Okay, so I need to solve the equation $10^x - 10 = 9990$. Let me think about how to approach this. Hmm, first, maybe I should isolate the term with the exponent. That usually helps with exponential equations.

So, starting with the original equation:

$10^x - 10 = 9990$

I can add 10 to both sides to get rid of the -10 on the left. Let me do that:

$10^x - 10 + 10 = 9990 + 10$

Simplifying both sides:

$10^x = 10000$

Alright, so now I have $10^x = 10000$. I need to solve for x. Since the base is 10, maybe I can express 10000 as a power of 10. Let me recall that $10^1 = 10$, $10^2 = 100$, $10^3 = 1000$, $10^4 = 10000$. Oh, right! So 10000 is $10^4$.

Therefore, substituting back into the equation:

$10^x = 10^4$

Since the bases are the same, the exponents must be equal. That means:

$x = 4$

Wait, let me check if that's correct. Let me substitute x = 4 back into the original equation to verify.

Original equation: $10^x - 10 = 9990$

Substituting x = 4:

Left side: $10^4 - 10 = 10000 - 10 = 9990$

Right side: 9990

Yes, both sides are equal. So x = 4 is indeed the solution.

But let me think again if there's another way to approach this problem, just to make sure I didn't miss anything. Maybe using logarithms? Let me try that method as well.

Starting again from $10^x = 10000$. Taking the logarithm base 10 of both sides:

$\log_{10}(10^x) = \log_{10}(10000)$

Simplifying the left side using logarithm properties: $\log_{10}(10^x) = x$, because $\log_b(b^k) = k$.

On the right side, $\log_{10}(10000)$. Since 10000 is $10^4$, this becomes $\log_{10}(10^4) = 4$.

Therefore, x = 4. Same result. So that confirms it again.

Alternatively, if I didn't recognize that 10000 is $10^4$, I could use natural logarithm or common logarithm. Let me try using natural logarithm for practice.

Starting from $10^x = 10000$. Taking natural logarithm on both sides:

$\ln(10^x) = \ln(10000)$

Using the power rule for logarithms: $x \ln(10) = \ln(10000)$

Therefore, solving for x:

$x = \frac{\ln(10000)}{\ln(10)}$

Now, calculating $\ln(10000)$. Since 10000 is $10^4$, $\ln(10^4) = 4 \ln(10)$. Therefore:

$x = \frac{4 \ln(10)}{\ln(10)} = 4$.

Same answer again. So regardless of the method, I get x = 4.

Is there any chance that there could be another solution? For example, sometimes with logarithms, you have to consider different branches or something? But in this case, since the original equation is $10^x = 10000$, and 10^x is a one-to-one function, there's only one solution. So x must be 4.

Let me also think about the graph of the function $f(x) = 10^x - 10$. It's an exponential growth function shifted down by 10 units. The equation $10^x - 10 = 9990$ is equivalent to finding the x-intercept of $f(x) - 9990 = 0$, which would be a single point since the function is strictly increasing. Therefore, only one solution exists, which is x = 4.

Another way to check is to consider the value of x. If x is 4, then 10^4 is 10000, which is exactly 10 more than 9990. So subtracting 10 gives 9990. Perfect.

Alternatively, if I thought maybe x is a decimal or fraction, but given that 10^x is 10000, which is a whole number, and since 10^x is 10 raised to some power, it's likely that x is an integer here.

Wait, but just to be thorough, suppose someone thought maybe x is something else. Let me try plugging in x = 5. Then 10^5 is 100000, subtract 10 is 99990, which is way bigger than 9990. So x=5 is too big. x=3: 10^3 is 1000, subtract 10 is 990, which is way smaller than 9990. So between x=3 and x=5, the value goes from 990 to 99990. Since 9990 is between those two, but we saw that x=4 gives exactly 9990. So that's the correct value.

Alternatively, if I didn't remember that 10^4 is 10000, maybe I could use logarithms. Let me try that again.

Original equation: $10^x = 10000$. Taking log base 10:

x = log10(10000). Since log10(10000) is the exponent you need to raise 10 to get 10000, which is 4.

Alternatively, if I didn't know that, I could use logarithm tables or a calculator, but since this is a math problem, it's expected to know that 10^4 is 10000.

Another thought: maybe the problem is designed to test understanding of exponents and logarithms. So, the key steps are isolating the exponential term, recognizing the power of 10, and then equating the exponents.

Alternatively, if someone is not comfortable with exponents, they might try to express 9990 + 10 = 10000, and then think about how many zeros are in 10000. 10000 has four zeros, so that's 10^4. Therefore, x is 4.

So, all methods lead to x = 4.

I think that's solid. I don't see any mistakes in my reasoning. All checks confirm that x = 4 is the correct solution.

**Final Answer**
The value of $ x $ is \boxed{4}.

To solve the equation:

$$
10^x - 10 = 9990
$$

we begin by isolating the exponential term. This is a standard approach for equations involving exponents.

---

### Step 1: Isolate the Exponential Term

Add 10 to both sides of the equation:

$$
10^x - 10 + 10 = 9990 + 10
$$

$$
10^x = 10000
$$

---

### Step 2: Express the Right-Hand Side as a Power of 10

We recognize that:

$$
10000 = 10^4
$$

So the equation becomes:

$$
10^x = 10^4
$$

---

### Step 3: Equate the Exponents

Since the bases are the same and the equation is in the form $ a^b = a^c $, we can equate the exponents:

$$
x = 4
$$

---

### Step 4: Verify the Solution

Substitute $ x = 4 $ back into the original equation:

$$
10^4 - 10 = 10000 - 10 = 9990
$$

This matches the right-hand side of the original equation, confirming that the solution is correct.

---

### Alternative Method: Using Logarithms

We can also solve the equation using logarithms. Starting from:

$$
10^x = 10000
$$

Take the logarithm base 10 of both sides:

$$
\log_{10}(10^x) = \log_{10}(10000)
$$

Using the logarithmic identity $ \log_b(b^k) = k $, we get:

$$
x = \log_{10}(10000)
$$

Since $ 10000 = 10^4 $, we have:

$$
x = \log_{10}(10^4) = 4
$$

---

### Conclusion

All methods confirm that the solution is:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation $10^x - 10 = 9990$. Let me think about how to approach this. Hmm, first, maybe I should isolate the term with the exponent. That usually helps with exponential equations.

So, starting with the original equation:

$10^x - 10 = 9990$

I can add 10 to both sides to get rid of the -10 on the left. Let me do that:

$10^x - 10 + 10 = 9990 + 10$

Simplifying both sides:

$10^x = 10000$

Alright, so now I have $10^x = 10000$. I need to solve for x. Since the base is 10, maybe I can express 10000 as a power of 10. Let me recall that $10^1 = 10$, $10^2 = 100$, $10^3 = 1000$, $10^4 = 10000$. Oh, right! So 10000 is $10^4$.

Therefore, substituting back into the equation:

$10^x = 10^4$

Since the bases are the same, the exponents must be equal. That means:

$x = 4$

Wait, let me check if that's correct. Let me substitute x = 4 back into the original equation to verify.

Original equation: $10^x - 10 = 9990$

Substituting x = 4:

Left side: $10^4 - 10 = 10000 - 10 = 9990$

Right side: 9990

Yes, both sides are equal. So x = 4 is indeed the solution.

But let me think again if there's another way to approach this problem, just to make sure I didn't miss anything. Maybe using logarithms? Let me try that method as well.

Starting again from $10^x = 10000$. Taking the logarithm base 10 of both sides:

$\log_{10}(10^x) = \log_{10}(10000)$

Simplifying the left side using logarithm properties: $\log_{10}(10^x) = x$, because $\log_b(b^k) = k$.

On the right side, $\log_{10}(10000)$. Since 10000 is $10^4$, this becomes $\log_{10}(10^4) = 4$.

Therefore, x = 4. Same result. So that confirms it again.

Alternatively, if I didn't recognize that 10000 is $10^4$, I could use natural logarithm or common logarithm. Let me try using natural logarithm for practice.

Starting from $10^x = 10000$. Taking natural logarithm on both sides:

$\ln(10^x) = \ln(10000)$

Using the power rule for logarithms: $x \ln(10) = \ln(10000)$

Therefore, solving for x:

$x = \frac{\ln(10000)}{\ln(10)}$

Now, calculating $\ln(10000)$. Since 10000 is $10^4$, $\ln(10^4) = 4 \ln(10)$. Therefore:

$x = \frac{4 \ln(10)}{\ln(10)} = 4$.

Same answer again. So regardless of the method, I get x = 4.

Is there any chance that there could be another solution? For example, sometimes with logarithms, you have to consider different branches or something? But in this case, since the original equation is $10^x = 10000$, and 10^x is a one-to-one function, there's only one solution. So x must be 4.

Let me also think about the graph of the function $f(x) = 10^x - 10$. It's an exponential growth function shifted down by 10 units. The equation $10^x - 10 = 9990$ is equivalent to finding the x-intercept of $f(x) - 9990 = 0$, which would be a single point since the function is strictly increasing. Therefore, only one solution exists, which is x = 4.

Another way to check is to consider the value of x. If x is 4, then 10^4 is 10000, which is exactly 10 more than 9990. So subtracting 10 gives 9990. Perfect.

Alternatively, if I thought maybe x is a decimal or fraction, but given that 10^x is 10000, which is a whole number, and since 10^x is 10 raised to some power, it's likely that x is an integer here.

Wait, but just to be thorough, suppose someone thought maybe x is something else. Let me try plugging in x = 5. Then 10^5 is 100000, subtract 10 is 99990, which is way bigger than 9990. So x=5 is too big. x=3: 10^3 is 1000, subtract 10 is 990, which is way smaller than 9990. So between x=3 and x=5, the value goes from 990 to 99990. Since 9990 is between those two, but we saw that x=4 gives exactly 9990. So that's the correct value.

Alternatively, if I didn't remember that 10^4 is 10000, maybe I could use logarithms. Let me try that again.

Original equation: $10^x = 10000$. Taking log base 10:

x = log10(10000). Since log10(10000) is the exponent you need to raise 10 to get 10000, which is 4.

Alternatively, if I didn't know that, I could use logarithm tables or a calculator, but since this is a math problem, it's expected to know that 10^4 is 10000.

Another thought: maybe the problem is designed to test understanding of exponents and logarithms. So, the key steps are isolating the exponential term, recognizing the power of 10, and then equating the exponents.

Alternatively, if someone is not comfortable with exponents, they might try to express 9990 + 10 = 10000, and then think about how many zeros are in 10000. 10000 has four zeros, so that's 10^4. Therefore, x is 4.

So, all methods lead to x = 4.

I think that's solid. I don't see any mistakes in my reasoning. All checks confirm that x = 4 is the correct solution.

**Final Answer**
The value of $ x $ is \boxed{4}.

To solve the equation:

$$
10^x - 10 = 9990
$$

we begin by isolating the exponential term. This is a standard approach for equations involving exponents.

---

### Step 1: Isolate the Exponential Term

Add 10 to both sides of the equation:

$$
10^x - 10 + 10 = 9990 + 10
$$

$$
10^x = 10000
$$

---

### Step 2: Express the Right-Hand Side as a Power of 10

We recognize that:

$$
10000 = 10^4
$$

So the equation becomes:

$$
10^x = 10^4
$$

---

### Step 3: Equate the Exponents

Since the bases are the same and the equation is in the form $ a^b = a^c $, we can equate the exponents:

$$
x = 4
$$

---

### Step 4: Verify the Solution

Substitute $ x = 4 $ back into the original equation:

$$
10^4 - 10 = 10000 - 10 = 9990
$$

This matches the right-hand side of the original equation, confirming that the solution is correct.

---

### Alternative Method: Using Logarithms

We can also solve the equation using logarithms. Starting from:

$$
10^x = 10000
$$

Take the logarithm base 10 of both sides:

$$
\log_{10}(10^x) = \log_{10}(10000)
$$

Using the logarithmic identity $ \log_b(b^k) = k $, we get:

$$
x = \log_{10}(10000)
$$

Since $ 10000 = 10^4 $, we have:

$$
x = \log_{10}(10^4) = 4
$$

---

### Conclusion

All methods confirm that the solution is:

$$
\boxed{4}
$$

Solution: Since $10^x - 10 = 9990,$ we have $$10^x = 9990+10=10000.$$If $10^x = 10000,$ then $x=\boxed{4},$ since $10000$ ends in four zeroes.


Explanation:
Okay, so I need to find the sum of the measures of angle J and angle H in this circle J diagram. Let me try to visualize the Asymptote figure they provided. There's a circle with center J, and points H, O, and N. HO and HN are tangent to the circle at O and N respectively. The diagram shows J connected to N and O, and then H connected to N and O, forming a quadrilateral JNHO? Wait, the Asymptote code draws J--N--H--O--cycle, so that's a quadrilateral with vertices J, N, H, O. So, the figure is a quadrilateral where two sides are tangents from H to the circle, and the other two sides are radii from J to N and O?

Wait, but tangents from a common external point are equal in length. So HO and HN are both tangents from H to the circle, so their lengths should be equal. Also, since HO and HN are tangent to the circle at O and N, then JO and JN are radii perpendicular to the tangents at the points of tangency. So, JO is perpendicular to HO, and JN is perpendicular to HN. That means angles JOH and JNH are right angles (90 degrees each).

So, quadrilateral JNHO has two right angles at O and N. Let me try to sketch this mentally. The circle is centered at J, with radius JO and JN. Points O and N are points of tangency from H. So, H is outside the circle, and HO and HN are tangent segments. Then, connecting J to N and J to O, forming two radii. Then connecting H to N and H to O. So quadrilateral JNHO has sides JO, ON, NH, and HJ? Wait, no. Wait, the Asymptote code draws J--N--H--O--cycle. So the quadrilateral is J connected to N, N connected to H, H connected to O, and O connected back to J. So the sides are JN, NH, HO, and OJ.

So, in this quadrilateral, we have two right angles at O and N. Because JO is perpendicular to HO, and JN is perpendicular to HN. Therefore, angles at O and N are 90 degrees. So, quadrilateral JNHO has two right angles. Now, the question is asking for the sum of the measures of angle J and angle H. So angle at J (which is angle OJN?) and angle at H (which is angle NHO?).

Wait, let me clarify. In quadrilateral JNHO, the vertices are J, N, H, O. So angle J is the angle at vertex J, which is between sides JN and JO. Similarly, angle H is the angle at vertex H, which is between sides HN and HO. So, we need to find m∠J + m∠H.

Since we know two angles of the quadrilateral (at O and N are 90 degrees each), maybe we can use the fact that the sum of the interior angles of a quadrilateral is 360 degrees. So, if we can find the measures of angles at J and H, then their sum would be 360 - 90 - 90 = 180 degrees? Wait, that would mean that m∠J + m∠H = 180 degrees? But that seems too straightforward. Is that correct?

Wait, let me check again. The sum of the interior angles of any quadrilateral is 360 degrees. If two of the angles are 90 degrees each, then the sum of the other two angles is 360 - 90 - 90 = 180 degrees. Therefore, m∠J + m∠H = 180 degrees. So is the answer 180? But maybe I need to verify this with another method to be sure.

Alternatively, maybe there's some properties of tangents and circles that can be used here. Let me think. Since HO and HN are tangents from H to the circle, then HO = HN. Therefore, triangle HON is an isosceles triangle with HO = HN. Also, JO and JN are radii, so JO = JN. Wait, but JO and JN are both radii of the circle, so they are equal. Therefore, triangle JON is also an isosceles triangle with JO = JN.

But how does that help with angles at J and H?

Alternatively, maybe we can consider triangles JHO and JHN? Wait, but H is outside the circle, and JO and JN are radii. Let me think about the angles. Since JO is perpendicular to HO, angle JOH is 90 degrees. Similarly, angle JNH is 90 degrees. So, if I consider triangle JHO, which has a right angle at O, and triangle JHN, which has a right angle at N.

But angle at J in triangle JHO is angle OJH, and angle at J in triangle JHN is angle NJH. Wait, but angle J in the quadrilateral is the angle between JN and JO, which is angle OJN. Wait, but if we consider triangle JHO and triangle JHN, then angle OJH and angle NJH are parts of angle J in the quadrilateral. Wait, maybe I need to think about the entire angle at J.

Wait, let me try to think of the quadrilateral again. The quadrilateral is JNHO with angles at J, N, H, O. We know angles at N and O are 90 degrees. So, if we can find angles at J and H, their sum is 180 degrees. But maybe there's another way to find them?

Alternatively, since HO and HN are tangents, then the line from H to J is the line connecting the external point to the center. In circle geometry, the line from the external point to the center bisects the angle between the two tangents. So, angle between HO and HN is bisected by HJ. Therefore, angle OHJ is equal to angle NJH.

But angle at H in the quadrilateral is angle NHO, which is different from the angle between the two tangents. Wait, maybe I need to consider triangle HJO and HJN. Let me try to find some relations.

Let me denote angle at J as angle OJN. Let me denote angle at H as angle NHO. Wait, but angle NHO is part of triangle HNO. Wait, maybe I need to use some trigonometry here. Let me assign some variables.

Let me denote the measure of angle J as x, and the measure of angle H as y. Then, according to the quadrilateral angle sum, x + y + 90 + 90 = 360, so x + y = 180 degrees. Therefore, the answer is 180 degrees. But is this correct? It seems straightforward, but maybe there's a trick here?

Wait, but maybe the quadrilateral isn't convex? Or maybe the angles at O and N are not both 90 degrees? Wait, the tangents are perpendicular to the radii at the points of contact. So, JO is perpendicular to HO, and JN is perpendicular to HN. Therefore, angles at O and N are right angles. Therefore, the quadrilateral has two right angles. Therefore, the sum of the other two angles is 180 degrees. Therefore, the answer is 180 degrees.

But let me check with another approach. Let me consider triangle JHO. In triangle JHO, we have a right angle at O. Similarly, triangle JHN has a right angle at N. Let me denote angle at J in triangle JHO as angle OJH, and angle at J in triangle JHN as angle NJH. Then, angle OJH + angle NJH = angle OJN, which is angle J in the quadrilateral.

Similarly, in triangle JHO, angle at H is angle JHO, and in triangle JHN, angle at H is angle JHN. Then, angle JHO + angle JHN = angle NHO, which is angle H in the quadrilateral.

But since HO and HN are both tangents from H, then triangles JHO and JHN are congruent? Because JO = JN (radii), HO = HN (tangents from same point), and HJ is common side. Therefore, triangles JHO and JHN are congruent by hypotenuse-leg theorem (right triangles with hypotenuse HJ and one leg equal). Therefore, angles OJH and NJH are equal, and angles JHO and JHN are equal.

Therefore, angle J in the quadrilateral is angle OJN = angle OJH + angle NJH = 2 * angle OJH. Similarly, angle H in the quadrilateral is angle NHO = angle JHO + angle JHN = 2 * angle JHO.

But in triangle JHO, angles sum to 180 degrees. So, angle at O is 90 degrees, angle at J is angle OJH, and angle at H is angle JHO. Therefore, angle OJH + angle JHO = 90 degrees. Similarly, angle JHO = 90 - angle OJH. Therefore, angle H in the quadrilateral is 2 * angle JHO = 2*(90 - angle OJH) = 180 - 2*angle OJH.

But angle J in the quadrilateral is 2*angle OJH. Therefore, angle J + angle H = 2*angle OJH + 180 - 2*angle OJH = 180 degrees.

Therefore, regardless of the specific measures, the sum is 180 degrees. Therefore, the answer is 180 degrees.

But let me check with coordinates. Maybe using coordinates can help. The Asymptote code defines J at (0,0), H at (6,0). Then, it defines a circle with center J (0,0) and radius 3. Then, it defines M as the midpoint of JH, which is (3,0). Then, it draws a circle with center M and radius 3. The intersection points of the two circles are O and N.

So, let me try to find coordinates of O and N. The first circle is centered at (0,0) with radius 3. The second circle is centered at (3,0) with radius 3. The intersection points of these two circles can be found by solving the equations:

First circle: x² + y² = 9

Second circle: (x - 3)² + y² = 9

Subtracting the first equation from the second:

(x - 3)² + y² - x² - y² = 9 - 9 => (x² -6x +9) - x² = 0 => -6x +9 = 0 => x = 9/6 = 3/2 = 1.5

So, x = 1.5. Plugging back into first equation:

(1.5)^2 + y^2 = 9 => 2.25 + y² = 9 => y² = 6.75 => y = ±√(27/4) = ±(3√3)/2 ≈ ±2.598

Therefore, the points of intersection are (1.5, (3√3)/2) and (1.5, -(3√3)/2). The Asymptote code labels O as tangentPoints[0] and N as tangentPoints[1]. Depending on the order, but since the Asymptote code draws J--N--H--O--cycle, probably N is the lower one and O is the upper one? Let me check.

Assuming that O is (1.5, (3√3)/2) and N is (1.5, -(3√3)/2). Then, let's compute angles at J and H.

First, coordinates:

J = (0,0)

H = (6,0)

O = (1.5, (3√3)/2)

N = (1.5, -(3√3)/2)

So, angle at J is angle OJN. Let me compute vectors JO and JN.

Vector JO is from J to O: (1.5, (3√3)/2)

Vector JN is from J to N: (1.5, -(3√3)/2)

So, angle between vectors JO and JN. Since JO and JN are symmetric with respect to the x-axis, the angle between them is twice the angle between JO and the x-axis.

Let me compute the angle of JO with the x-axis. The slope of JO is [(3√3)/2]/1.5 = [(3√3)/2]/(3/2) = √3. Therefore, the angle is arctangent(√3) which is 60 degrees. Therefore, the angle between JO and the x-axis is 60 degrees, and between JN and the x-axis is -60 degrees. Therefore, the angle between JO and JN is 60 - (-60) = 120 degrees. Therefore, angle J is 120 degrees.

Now, angle at H is angle NHO. Let me compute that. Points N, H, O.

Point H is at (6,0), N is at (1.5, -(3√3)/2), O is at (1.5, (3√3)/2). So, angle at H between points N, H, O.

To compute angle NHO, we can compute the angle between vectors HN and HO.

First, compute vectors HN and HO.

Vector HN is N - H = (1.5 - 6, -(3√3)/2 - 0) = (-4.5, -(3√3)/2)

Vector HO is O - H = (1.5 - 6, (3√3)/2 - 0) = (-4.5, (3√3)/2)

So, vectors HN = (-4.5, -(3√3)/2) and HO = (-4.5, (3√3)/2). The angle between these two vectors can be found using the dot product formula.

Dot product of HN and HO is (-4.5)(-4.5) + [-(3√3)/2][(3√3)/2] = 20.25 + [ - (9*3)/4 ] = 20.25 - 27/4 = 20.25 - 6.75 = 13.5

The magnitude of HN is sqrt[(-4.5)^2 + (-(3√3)/2)^2] = sqrt[20.25 + (27/4)] = sqrt[20.25 + 6.75] = sqrt[27] = 3√3

Similarly, magnitude of HO is the same, 3√3.

Therefore, the cosine of the angle between HN and HO is (13.5)/(3√3 * 3√3) = 13.5/(9*3) = 13.5/27 = 0.5

Therefore, the angle is arccos(0.5) = 60 degrees. Therefore, angle H is 60 degrees.

Therefore, angle J is 120 degrees, angle H is 60 degrees, so their sum is 120 + 60 = 180 degrees.

So, this confirms the previous conclusion. Therefore, the answer is 180 degrees.

But let me check once more. In the coordinate system, angle J is 120 degrees, angle H is 60 degrees. Their sum is 180. So, regardless of the specific configuration, the sum is always 180 degrees. Therefore, the answer is 180 degrees.

Alternatively, since angle J is the angle between two radii JO and JN, which are separated by some angle. In the coordinate system, the angle between JO and JN was 120 degrees. But in general, for any circle with two tangents from an external point, the angle between the two radii JO and JN is related to the angle between the tangents.

In general, the angle between the two tangents from an external point H is equal to the difference between 180 degrees and the central angle between the two points of tangency. Wait, let me recall the formula.

If θ is the angle between the two tangents from H, then θ = 2 arcsin(r/d), where r is the radius and d is the distance from H to the center. Alternatively, the central angle between the two points of tangency is 2 arcsin(r/d). Wait, but in our case, we found that angle between the two radii JO and JN is 120 degrees, and the angle between the two tangents at H is 60 degrees. Indeed, 120 + 60 = 180. So, in general, the angle between the two tangents plus the central angle between the points of tangency equals 180 degrees.

Therefore, in this problem, angle at H (the angle between the two tangents) plus the central angle (angle J) equals 180 degrees. Therefore, their sum is 180 degrees.

Therefore, regardless of the specific configuration, the answer is 180 degrees.

So, after multiple approaches—using quadrilateral angle sum, coordinate geometry, and general circle tangent properties—it seems that the sum of the measures of angle J and angle H is indeed 180 degrees.

**Final Answer**
\boxed{180}

To solve the problem, we analyze the geometric configuration involving circle $ J $, with two tangent lines $ HO $ and $ HN $ drawn from an external point $ H $ to the circle, touching it at points $ O $ and $ N $, respectively.

---

### Step 1: Understand the Geometry

- **Circle Center**: $ J $ is the center of the circle.
- **Tangents**: $ HO $ and $ HN $ are tangents from point $ H $ to the circle, touching it at points $ O $ and $ N $, respectively.
- **Key Property**: A tangent to a circle is perpendicular to the radius at the point of tangency. Therefore:
  - $ \angle JOH = 90^\circ $
  - $ \angle JNH = 90^\circ $

This means that the quadrilateral $ JNHO $ has two right angles at $ O $ and $ N $.

---

### Step 2: Use the Sum of Interior Angles of a Quadrilateral

The sum of the interior angles of any quadrilateral is always:

$$
\text{Sum of angles} = 360^\circ
$$

We are given that:

- $ \angle O = 90^\circ $
- $ \angle N = 90^\circ $

Let $ \angle J $ and $ \angle H $ be the remaining two angles of the quadrilateral. Then:

$$
\angle J + \angle H + 90^\circ + 90^\circ = 360^\circ
$$

$$
\angle J + \angle H = 360^\circ - 180^\circ = 180^\circ
$$

---

### Step 3: Confirm with Coordinate Geometry (Optional Verification)

Using the Asymptote code, we can assign coordinates:

- $ J = (0, 0) $
- $ H = (6, 0) $
- The circle has radius 3, centered at $ J $, and the second circle centered at the midpoint of $ JH $, which is $ (3, 0) $, also with radius 3.

Solving the system of equations for the intersection points of the two circles gives the coordinates of $ O $ and $ N $, and computing the angles at $ J $ and $ H $ using vector analysis confirms that:

- $ \angle J = 120^\circ $
- $ \angle H = 60^\circ $

Their sum is:

$$
120^\circ + 60^\circ = 180^\circ
$$

---

### Final Conclusion

Regardless of the specific configuration, the sum of the measures of $ \angle J $ and $ \angle H $ is always:

$$
\boxed{180}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of the measures of angle J and angle H in this circle J diagram. Let me try to visualize the Asymptote figure they provided. There's a circle with center J, and points H, O, and N. HO and HN are tangent to the circle at O and N respectively. The diagram shows J connected to N and O, and then H connected to N and O, forming a quadrilateral JNHO? Wait, the Asymptote code draws J--N--H--O--cycle, so that's a quadrilateral with vertices J, N, H, O. So, the figure is a quadrilateral where two sides are tangents from H to the circle, and the other two sides are radii from J to N and O?

Wait, but tangents from a common external point are equal in length. So HO and HN are both tangents from H to the circle, so their lengths should be equal. Also, since HO and HN are tangent to the circle at O and N, then JO and JN are radii perpendicular to the tangents at the points of tangency. So, JO is perpendicular to HO, and JN is perpendicular to HN. That means angles JOH and JNH are right angles (90 degrees each).

So, quadrilateral JNHO has two right angles at O and N. Let me try to sketch this mentally. The circle is centered at J, with radius JO and JN. Points O and N are points of tangency from H. So, H is outside the circle, and HO and HN are tangent segments. Then, connecting J to N and J to O, forming two radii. Then connecting H to N and H to O. So quadrilateral JNHO has sides JO, ON, NH, and HJ? Wait, no. Wait, the Asymptote code draws J--N--H--O--cycle. So the quadrilateral is J connected to N, N connected to H, H connected to O, and O connected back to J. So the sides are JN, NH, HO, and OJ.

So, in this quadrilateral, we have two right angles at O and N. Because JO is perpendicular to HO, and JN is perpendicular to HN. Therefore, angles at O and N are 90 degrees. So, quadrilateral JNHO has two right angles. Now, the question is asking for the sum of the measures of angle J and angle H. So angle at J (which is angle OJN?) and angle at H (which is angle NHO?).

Wait, let me clarify. In quadrilateral JNHO, the vertices are J, N, H, O. So angle J is the angle at vertex J, which is between sides JN and JO. Similarly, angle H is the angle at vertex H, which is between sides HN and HO. So, we need to find m∠J + m∠H.

Since we know two angles of the quadrilateral (at O and N are 90 degrees each), maybe we can use the fact that the sum of the interior angles of a quadrilateral is 360 degrees. So, if we can find the measures of angles at J and H, then their sum would be 360 - 90 - 90 = 180 degrees? Wait, that would mean that m∠J + m∠H = 180 degrees? But that seems too straightforward. Is that correct?

Wait, let me check again. The sum of the interior angles of any quadrilateral is 360 degrees. If two of the angles are 90 degrees each, then the sum of the other two angles is 360 - 90 - 90 = 180 degrees. Therefore, m∠J + m∠H = 180 degrees. So is the answer 180? But maybe I need to verify this with another method to be sure.

Alternatively, maybe there's some properties of tangents and circles that can be used here. Let me think. Since HO and HN are tangents from H to the circle, then HO = HN. Therefore, triangle HON is an isosceles triangle with HO = HN. Also, JO and JN are radii, so JO = JN. Wait, but JO and JN are both radii of the circle, so they are equal. Therefore, triangle JON is also an isosceles triangle with JO = JN.

But how does that help with angles at J and H?

Alternatively, maybe we can consider triangles JHO and JHN? Wait, but H is outside the circle, and JO and JN are radii. Let me think about the angles. Since JO is perpendicular to HO, angle JOH is 90 degrees. Similarly, angle JNH is 90 degrees. So, if I consider triangle JHO, which has a right angle at O, and triangle JHN, which has a right angle at N.

But angle at J in triangle JHO is angle OJH, and angle at J in triangle JHN is angle NJH. Wait, but angle J in the quadrilateral is the angle between JN and JO, which is angle OJN. Wait, but if we consider triangle JHO and triangle JHN, then angle OJH and angle NJH are parts of angle J in the quadrilateral. Wait, maybe I need to think about the entire angle at J.

Wait, let me try to think of the quadrilateral again. The quadrilateral is JNHO with angles at J, N, H, O. We know angles at N and O are 90 degrees. So, if we can find angles at J and H, their sum is 180 degrees. But maybe there's another way to find them?

Alternatively, since HO and HN are tangents, then the line from H to J is the line connecting the external point to the center. In circle geometry, the line from the external point to the center bisects the angle between the two tangents. So, angle between HO and HN is bisected by HJ. Therefore, angle OHJ is equal to angle NJH.

But angle at H in the quadrilateral is angle NHO, which is different from the angle between the two tangents. Wait, maybe I need to consider triangle HJO and HJN. Let me try to find some relations.

Let me denote angle at J as angle OJN. Let me denote angle at H as angle NHO. Wait, but angle NHO is part of triangle HNO. Wait, maybe I need to use some trigonometry here. Let me assign some variables.

Let me denote the measure of angle J as x, and the measure of angle H as y. Then, according to the quadrilateral angle sum, x + y + 90 + 90 = 360, so x + y = 180 degrees. Therefore, the answer is 180 degrees. But is this correct? It seems straightforward, but maybe there's a trick here?

Wait, but maybe the quadrilateral isn't convex? Or maybe the angles at O and N are not both 90 degrees? Wait, the tangents are perpendicular to the radii at the points of contact. So, JO is perpendicular to HO, and JN is perpendicular to HN. Therefore, angles at O and N are right angles. Therefore, the quadrilateral has two right angles. Therefore, the sum of the other two angles is 180 degrees. Therefore, the answer is 180 degrees.

But let me check with another approach. Let me consider triangle JHO. In triangle JHO, we have a right angle at O. Similarly, triangle JHN has a right angle at N. Let me denote angle at J in triangle JHO as angle OJH, and angle at J in triangle JHN as angle NJH. Then, angle OJH + angle NJH = angle OJN, which is angle J in the quadrilateral.

Similarly, in triangle JHO, angle at H is angle JHO, and in triangle JHN, angle at H is angle JHN. Then, angle JHO + angle JHN = angle NHO, which is angle H in the quadrilateral.

But since HO and HN are both tangents from H, then triangles JHO and JHN are congruent? Because JO = JN (radii), HO = HN (tangents from same point), and HJ is common side. Therefore, triangles JHO and JHN are congruent by hypotenuse-leg theorem (right triangles with hypotenuse HJ and one leg equal). Therefore, angles OJH and NJH are equal, and angles JHO and JHN are equal.

Therefore, angle J in the quadrilateral is angle OJN = angle OJH + angle NJH = 2 * angle OJH. Similarly, angle H in the quadrilateral is angle NHO = angle JHO + angle JHN = 2 * angle JHO.

But in triangle JHO, angles sum to 180 degrees. So, angle at O is 90 degrees, angle at J is angle OJH, and angle at H is angle JHO. Therefore, angle OJH + angle JHO = 90 degrees. Similarly, angle JHO = 90 - angle OJH. Therefore, angle H in the quadrilateral is 2 * angle JHO = 2*(90 - angle OJH) = 180 - 2*angle OJH.

But angle J in the quadrilateral is 2*angle OJH. Therefore, angle J + angle H = 2*angle OJH + 180 - 2*angle OJH = 180 degrees.

Therefore, regardless of the specific measures, the sum is 180 degrees. Therefore, the answer is 180 degrees.

But let me check with coordinates. Maybe using coordinates can help. The Asymptote code defines J at (0,0), H at (6,0). Then, it defines a circle with center J (0,0) and radius 3. Then, it defines M as the midpoint of JH, which is (3,0). Then, it draws a circle with center M and radius 3. The intersection points of the two circles are O and N.

So, let me try to find coordinates of O and N. The first circle is centered at (0,0) with radius 3. The second circle is centered at (3,0) with radius 3. The intersection points of these two circles can be found by solving the equations:

First circle: x² + y² = 9

Second circle: (x - 3)² + y² = 9

Subtracting the first equation from the second:

(x - 3)² + y² - x² - y² = 9 - 9 => (x² -6x +9) - x² = 0 => -6x +9 = 0 => x = 9/6 = 3/2 = 1.5

So, x = 1.5. Plugging back into first equation:

(1.5)^2 + y^2 = 9 => 2.25 + y² = 9 => y² = 6.75 => y = ±√(27/4) = ±(3√3)/2 ≈ ±2.598

Therefore, the points of intersection are (1.5, (3√3)/2) and (1.5, -(3√3)/2). The Asymptote code labels O as tangentPoints[0] and N as tangentPoints[1]. Depending on the order, but since the Asymptote code draws J--N--H--O--cycle, probably N is the lower one and O is the upper one? Let me check.

Assuming that O is (1.5, (3√3)/2) and N is (1.5, -(3√3)/2). Then, let's compute angles at J and H.

First, coordinates:

J = (0,0)

H = (6,0)

O = (1.5, (3√3)/2)

N = (1.5, -(3√3)/2)

So, angle at J is angle OJN. Let me compute vectors JO and JN.

Vector JO is from J to O: (1.5, (3√3)/2)

Vector JN is from J to N: (1.5, -(3√3)/2)

So, angle between vectors JO and JN. Since JO and JN are symmetric with respect to the x-axis, the angle between them is twice the angle between JO and the x-axis.

Let me compute the angle of JO with the x-axis. The slope of JO is [(3√3)/2]/1.5 = [(3√3)/2]/(3/2) = √3. Therefore, the angle is arctangent(√3) which is 60 degrees. Therefore, the angle between JO and the x-axis is 60 degrees, and between JN and the x-axis is -60 degrees. Therefore, the angle between JO and JN is 60 - (-60) = 120 degrees. Therefore, angle J is 120 degrees.

Now, angle at H is angle NHO. Let me compute that. Points N, H, O.

Point H is at (6,0), N is at (1.5, -(3√3)/2), O is at (1.5, (3√3)/2). So, angle at H between points N, H, O.

To compute angle NHO, we can compute the angle between vectors HN and HO.

First, compute vectors HN and HO.

Vector HN is N - H = (1.5 - 6, -(3√3)/2 - 0) = (-4.5, -(3√3)/2)

Vector HO is O - H = (1.5 - 6, (3√3)/2 - 0) = (-4.5, (3√3)/2)

So, vectors HN = (-4.5, -(3√3)/2) and HO = (-4.5, (3√3)/2). The angle between these two vectors can be found using the dot product formula.

Dot product of HN and HO is (-4.5)(-4.5) + [-(3√3)/2][(3√3)/2] = 20.25 + [ - (9*3)/4 ] = 20.25 - 27/4 = 20.25 - 6.75 = 13.5

The magnitude of HN is sqrt[(-4.5)^2 + (-(3√3)/2)^2] = sqrt[20.25 + (27/4)] = sqrt[20.25 + 6.75] = sqrt[27] = 3√3

Similarly, magnitude of HO is the same, 3√3.

Therefore, the cosine of the angle between HN and HO is (13.5)/(3√3 * 3√3) = 13.5/(9*3) = 13.5/27 = 0.5

Therefore, the angle is arccos(0.5) = 60 degrees. Therefore, angle H is 60 degrees.

Therefore, angle J is 120 degrees, angle H is 60 degrees, so their sum is 120 + 60 = 180 degrees.

So, this confirms the previous conclusion. Therefore, the answer is 180 degrees.

But let me check once more. In the coordinate system, angle J is 120 degrees, angle H is 60 degrees. Their sum is 180. So, regardless of the specific configuration, the sum is always 180 degrees. Therefore, the answer is 180 degrees.

Alternatively, since angle J is the angle between two radii JO and JN, which are separated by some angle. In the coordinate system, the angle between JO and JN was 120 degrees. But in general, for any circle with two tangents from an external point, the angle between the two radii JO and JN is related to the angle between the tangents.

In general, the angle between the two tangents from an external point H is equal to the difference between 180 degrees and the central angle between the two points of tangency. Wait, let me recall the formula.

If θ is the angle between the two tangents from H, then θ = 2 arcsin(r/d), where r is the radius and d is the distance from H to the center. Alternatively, the central angle between the two points of tangency is 2 arcsin(r/d). Wait, but in our case, we found that angle between the two radii JO and JN is 120 degrees, and the angle between the two tangents at H is 60 degrees. Indeed, 120 + 60 = 180. So, in general, the angle between the two tangents plus the central angle between the points of tangency equals 180 degrees.

Therefore, in this problem, angle at H (the angle between the two tangents) plus the central angle (angle J) equals 180 degrees. Therefore, their sum is 180 degrees.

Therefore, regardless of the specific configuration, the answer is 180 degrees.

So, after multiple approaches—using quadrilateral angle sum, coordinate geometry, and general circle tangent properties—it seems that the sum of the measures of angle J and angle H is indeed 180 degrees.

**Final Answer**
\boxed{180}

To solve the problem, we analyze the geometric configuration involving circle $ J $, with two tangent lines $ HO $ and $ HN $ drawn from an external point $ H $ to the circle, touching it at points $ O $ and $ N $, respectively.

---

### Step 1: Understand the Geometry

- **Circle Center**: $ J $ is the center of the circle.
- **Tangents**: $ HO $ and $ HN $ are tangents from point $ H $ to the circle, touching it at points $ O $ and $ N $, respectively.
- **Key Property**: A tangent to a circle is perpendicular to the radius at the point of tangency. Therefore:
  - $ \angle JOH = 90^\circ $
  - $ \angle JNH = 90^\circ $

This means that the quadrilateral $ JNHO $ has two right angles at $ O $ and $ N $.

---

### Step 2: Use the Sum of Interior Angles of a Quadrilateral

The sum of the interior angles of any quadrilateral is always:

$$
\text{Sum of angles} = 360^\circ
$$

We are given that:

- $ \angle O = 90^\circ $
- $ \angle N = 90^\circ $

Let $ \angle J $ and $ \angle H $ be the remaining two angles of the quadrilateral. Then:

$$
\angle J + \angle H + 90^\circ + 90^\circ = 360^\circ
$$

$$
\angle J + \angle H = 360^\circ - 180^\circ = 180^\circ
$$

---

### Step 3: Confirm with Coordinate Geometry (Optional Verification)

Using the Asymptote code, we can assign coordinates:

- $ J = (0, 0) $
- $ H = (6, 0) $
- The circle has radius 3, centered at $ J $, and the second circle centered at the midpoint of $ JH $, which is $ (3, 0) $, also with radius 3.

Solving the system of equations for the intersection points of the two circles gives the coordinates of $ O $ and $ N $, and computing the angles at $ J $ and $ H $ using vector analysis confirms that:

- $ \angle J = 120^\circ $
- $ \angle H = 60^\circ $

Their sum is:

$$
120^\circ + 60^\circ = 180^\circ
$$

---

### Final Conclusion

Regardless of the specific configuration, the sum of the measures of $ \angle J $ and $ \angle H $ is always:

$$
\boxed{180}
$$

Solution: Since $\overline{OH}$ and $\overline{NH}$ are tangent to radii of the circle at $O$ and $N$, we have $\angle O =\angle N = 90^\circ$.  The sum of the measures of the interior angles of quadrilateral $JOHN$ is $360^\circ$, so $\angle J + \angle H = 360^\circ - \angle O - \angle N = \boxed{180^\circ}$.