Explanation:
Okay, so I need to find the units digit of 18 raised to the 6th power. Hmm, let me think. The units digit is the last digit of the number, right? So, if I can figure out what 18^6 ends with, that's the answer.

First, maybe I should recall that when you raise a number to a power, the units digit depends only on the units digit of the original number. Wait, is that true? Let me check. For example, if I have 18, the units digit is 8. So maybe I can just focus on 8^6 and find its units digit? That might be easier. Let me verify this idea with a smaller exponent. Let's take 18^1, which is 18, units digit 8. 18^2 is 324, units digit 4. If I do 8^2, that's 64, which also ends with 4. Okay, so that seems to work. Similarly, 18^3 would be 18*18*18. Let me compute 18*18 first, which is 324, then 324*18. Let me just focus on the units digit. 324 ends with 4, so 4*8 is 32, so the units digit is 2. If I do 8^3, that's 512, which also ends with 2. So yeah, it seems like the units digit of 18^n is the same as the units digit of 8^n. Therefore, I can simplify the problem to finding the units digit of 8^6.

Alright, so now I need to compute 8^6 and find its units digit. Let me compute 8^1 through 8^6 step by step and note the units digit each time.

Starting with 8^1: 8. Units digit is 8.

8^2: 8*8 = 64. Units digit is 4.

8^3: 8^2 * 8 = 64*8. Let me compute 64*8. 60*8=480, 4*8=32, so total is 480+32=512. Units digit is 2.

8^4: 8^3 * 8 = 512*8. Let me compute that. 500*8=4000, 12*8=96, so total is 4000+96=4096. Units digit is 6.

8^5: 8^4 * 8 = 4096*8. Let me do 4000*8=32000, 96*8=768. Adding them together: 32000 + 768 = 32768. Units digit is 8.

Wait, 8^5 ends with 8? Let me check that again. 8^4 is 4096, which ends with 6. 6*8 is 48, so units digit 8. Yes, that's correct. So 8^5 ends with 8.

Then 8^6 would be 8^5 * 8. So, units digit is 8*8=64, so units digit 4.

Therefore, the units digit of 8^6 is 4, so the units digit of 18^6 is also 4.

But let me verify this another way, maybe by looking for a pattern or using modular arithmetic.

Since we're only interested in the units digit, we can compute everything modulo 10. So, 18 ≡ 8 mod 10. Therefore, 18^6 ≡ 8^6 mod 10.

So, if I can compute 8^6 mod 10, that's the answer.

Alternatively, maybe there's a cycle in the units digits of powers of 8. Let me check.

From earlier calculations:

8^1 mod 10 = 8

8^2 mod 10 = 4

8^3 mod 10 = 2

8^4 mod 10 = 6

8^5 mod 10 = 8

8^6 mod 10 = 4

So, the cycle here seems to be 8, 4, 2, 6, and then repeats every 4 exponents. Let me check:

8^1: 8

8^2: 4

8^3: 2

8^4: 6

8^5: 8 (same as 8^1)

8^6: 4 (same as 8^2)

So the cycle length is 4. Therefore, the pattern repeats every 4 powers.

Therefore, to find 8^6 mod 10, we can divide the exponent by 4 and find the remainder.

So, 6 divided by 4 is 1 with a remainder of 2. Therefore, 8^6 mod 10 is the same as 8^2 mod 10, which is 4.

Therefore, the units digit is 4.

Alternatively, if I didn't notice the cycle, I could compute 8^6 step by step:

8^1 = 8

8^2 = 64

8^3 = 512

8^4 = 4096

8^5 = 32768

8^6 = 262144

So, 8^6 is 262144, which ends with 4.

Therefore, the units digit is 4.

But let me check if I did that exponentiation correctly. Let me compute 8^4: 8^3 is 512, times 8 is 4096. Correct. Then 8^5 is 4096*8. Let me compute 4096*8:

Well, 4000*8 = 32,000

96*8 = 768

Adding them gives 32,000 + 768 = 32,768. Correct. Then 8^6 is 32,768 * 8. Let me compute that:

32,768 * 8:

First, 30,000 * 8 = 240,000

2,000 * 8 = 16,000

700 * 8 = 5,600

60 * 8 = 480

8 * 8 = 64

Adding them all up:

240,000 + 16,000 = 256,000

256,000 + 5,600 = 261,600

261,600 + 480 = 262,080

262,080 + 64 = 262,144.

Yes, so 8^6 is 262,144, which indeed ends with 4.

Therefore, the units digit is 4.

Alternatively, maybe I can use Euler's theorem or something? Wait, Euler's theorem says that if two numbers are coprime, then a^φ(n) ≡ 1 mod n. Here, n is 10, φ(10) = 4. Since 8 and 10 are not coprime (they share a common factor of 2), Euler's theorem doesn't apply directly. So maybe that's not helpful here.

But if I consider modulo 10, and since 8 and 10 are not coprime, perhaps another approach. But since we already found the pattern, maybe that's sufficient.

Alternatively, maybe I can use the fact that 8^4 = 4096 ≡ 6 mod 10. Then 8^4 ≡ 6 mod 10. Then 8^5 = 8^4 * 8 ≡ 6*8 = 48 ≡ 8 mod 10. Then 8^6 = 8^5 * 8 ≡ 8*8 = 64 ≡ 4 mod 10. So that's another way to see it.

So, regardless of the method, the units digit is 4.

Therefore, the answer is 4.

But just to make sure I didn't make any mistakes in my reasoning, let me check once again.

Original problem: units digit of 18^6.

Since 18 ends with 8, the units digit of 18^6 is same as units digit of 8^6.

Calculating 8^1 to 8^6:

8^1 = 8

8^2 = 64

8^3 = 512

8^4 = 4096

8^5 = 32768

8^6 = 262144

Units digits: 8, 4, 2, 6, 8, 4.

So, yes, the pattern cycles every 4 exponents: 8, 4, 2, 6, then repeats.

So, exponent 1: 8

Exponent 2: 4

Exponent 3: 2

Exponent 4: 6

Exponent 5: 8 (same as exponent 1)

Exponent 6: 4 (same as exponent 2)

Therefore, exponent 6 corresponds to the second position in the cycle, which is 4.

Therefore, the units digit is 4.

Alternatively, since the cycle length is 4, we can compute 6 divided by 4, which gives a remainder of 2. Therefore, the units digit is the same as 8^2, which is 4.

All methods lead to the same conclusion.

Therefore, I can be confident that the units digit of 18^6 is 4.

**Final Answer**
The units digit of $18^6$ is \boxed{4}.

To determine the **units digit** of $ 18^6 $, we focus on the **last digit** of the result, which is determined only by the **units digit** of the base number raised to the given power.

---

### Step 1: Simplify the Problem

The number $ 18 $ has a **units digit of 8**, so we can simplify the problem to finding the **units digit of $ 8^6 $**.

This is because the **units digit of a power** depends only on the **units digit of the base**. For example:

- $ 18^2 = 324 $ → units digit is 4
- $ 8^2 = 64 $ → units digit is 4

So, we can focus on computing the units digit of $ 8^6 $.

---

### Step 2: Compute Powers of 8 and Observe the Pattern

Let’s compute the units digits of successive powers of 8:

- $ 8^1 = 8 $ → units digit = **8**
- $ 8^2 = 64 $ → units digit = **4**
- $ 8^3 = 512 $ → units digit = **2**
- $ 8^4 = 4096 $ → units digit = **6**
- $ 8^5 = 32768 $ → units digit = **8**
- $ 8^6 = 262144 $ → units digit = **4**

We observe a **repeating cycle** of units digits:
**8 → 4 → 2 → 6 → 8 → 4 → ...**

This cycle has a **length of 4**.

---

### Step 3: Use the Pattern to Find the Units Digit of $ 8^6 $

Since the cycle repeats every 4 exponents, we can find the position of the exponent 6 in the cycle by computing:

$$
6 \mod 4 = 2
$$

This means the units digit of $ 8^6 $ is the **same as the units digit of $ 8^2 $**, which is **4**.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the units digit of 18 raised to the 6th power. Hmm, let me think. The units digit is the last digit of the number, right? So, if I can figure out what 18^6 ends with, that's the answer.

First, maybe I should recall that when you raise a number to a power, the units digit depends only on the units digit of the original number. Wait, is that true? Let me check. For example, if I have 18, the units digit is 8. So maybe I can just focus on 8^6 and find its units digit? That might be easier. Let me verify this idea with a smaller exponent. Let's take 18^1, which is 18, units digit 8. 18^2 is 324, units digit 4. If I do 8^2, that's 64, which also ends with 4. Okay, so that seems to work. Similarly, 18^3 would be 18*18*18. Let me compute 18*18 first, which is 324, then 324*18. Let me just focus on the units digit. 324 ends with 4, so 4*8 is 32, so the units digit is 2. If I do 8^3, that's 512, which also ends with 2. So yeah, it seems like the units digit of 18^n is the same as the units digit of 8^n. Therefore, I can simplify the problem to finding the units digit of 8^6.

Alright, so now I need to compute 8^6 and find its units digit. Let me compute 8^1 through 8^6 step by step and note the units digit each time.

Starting with 8^1: 8. Units digit is 8.

8^2: 8*8 = 64. Units digit is 4.

8^3: 8^2 * 8 = 64*8. Let me compute 64*8. 60*8=480, 4*8=32, so total is 480+32=512. Units digit is 2.

8^4: 8^3 * 8 = 512*8. Let me compute that. 500*8=4000, 12*8=96, so total is 4000+96=4096. Units digit is 6.

8^5: 8^4 * 8 = 4096*8. Let me do 4000*8=32000, 96*8=768. Adding them together: 32000 + 768 = 32768. Units digit is 8.

Wait, 8^5 ends with 8? Let me check that again. 8^4 is 4096, which ends with 6. 6*8 is 48, so units digit 8. Yes, that's correct. So 8^5 ends with 8.

Then 8^6 would be 8^5 * 8. So, units digit is 8*8=64, so units digit 4.

Therefore, the units digit of 8^6 is 4, so the units digit of 18^6 is also 4.

But let me verify this another way, maybe by looking for a pattern or using modular arithmetic.

Since we're only interested in the units digit, we can compute everything modulo 10. So, 18 ≡ 8 mod 10. Therefore, 18^6 ≡ 8^6 mod 10.

So, if I can compute 8^6 mod 10, that's the answer.

Alternatively, maybe there's a cycle in the units digits of powers of 8. Let me check.

From earlier calculations:

8^1 mod 10 = 8

8^2 mod 10 = 4

8^3 mod 10 = 2

8^4 mod 10 = 6

8^5 mod 10 = 8

8^6 mod 10 = 4

So, the cycle here seems to be 8, 4, 2, 6, and then repeats every 4 exponents. Let me check:

8^1: 8

8^2: 4

8^3: 2

8^4: 6

8^5: 8 (same as 8^1)

8^6: 4 (same as 8^2)

So the cycle length is 4. Therefore, the pattern repeats every 4 powers.

Therefore, to find 8^6 mod 10, we can divide the exponent by 4 and find the remainder.

So, 6 divided by 4 is 1 with a remainder of 2. Therefore, 8^6 mod 10 is the same as 8^2 mod 10, which is 4.

Therefore, the units digit is 4.

Alternatively, if I didn't notice the cycle, I could compute 8^6 step by step:

8^1 = 8

8^2 = 64

8^3 = 512

8^4 = 4096

8^5 = 32768

8^6 = 262144

So, 8^6 is 262144, which ends with 4.

Therefore, the units digit is 4.

But let me check if I did that exponentiation correctly. Let me compute 8^4: 8^3 is 512, times 8 is 4096. Correct. Then 8^5 is 4096*8. Let me compute 4096*8:

Well, 4000*8 = 32,000

96*8 = 768

Adding them gives 32,000 + 768 = 32,768. Correct. Then 8^6 is 32,768 * 8. Let me compute that:

32,768 * 8:

First, 30,000 * 8 = 240,000

2,000 * 8 = 16,000

700 * 8 = 5,600

60 * 8 = 480

8 * 8 = 64

Adding them all up:

240,000 + 16,000 = 256,000

256,000 + 5,600 = 261,600

261,600 + 480 = 262,080

262,080 + 64 = 262,144.

Yes, so 8^6 is 262,144, which indeed ends with 4.

Therefore, the units digit is 4.

Alternatively, maybe I can use Euler's theorem or something? Wait, Euler's theorem says that if two numbers are coprime, then a^φ(n) ≡ 1 mod n. Here, n is 10, φ(10) = 4. Since 8 and 10 are not coprime (they share a common factor of 2), Euler's theorem doesn't apply directly. So maybe that's not helpful here.

But if I consider modulo 10, and since 8 and 10 are not coprime, perhaps another approach. But since we already found the pattern, maybe that's sufficient.

Alternatively, maybe I can use the fact that 8^4 = 4096 ≡ 6 mod 10. Then 8^4 ≡ 6 mod 10. Then 8^5 = 8^4 * 8 ≡ 6*8 = 48 ≡ 8 mod 10. Then 8^6 = 8^5 * 8 ≡ 8*8 = 64 ≡ 4 mod 10. So that's another way to see it.

So, regardless of the method, the units digit is 4.

Therefore, the answer is 4.

But just to make sure I didn't make any mistakes in my reasoning, let me check once again.

Original problem: units digit of 18^6.

Since 18 ends with 8, the units digit of 18^6 is same as units digit of 8^6.

Calculating 8^1 to 8^6:

8^1 = 8

8^2 = 64

8^3 = 512

8^4 = 4096

8^5 = 32768

8^6 = 262144

Units digits: 8, 4, 2, 6, 8, 4.

So, yes, the pattern cycles every 4 exponents: 8, 4, 2, 6, then repeats.

So, exponent 1: 8

Exponent 2: 4

Exponent 3: 2

Exponent 4: 6

Exponent 5: 8 (same as exponent 1)

Exponent 6: 4 (same as exponent 2)

Therefore, exponent 6 corresponds to the second position in the cycle, which is 4.

Therefore, the units digit is 4.

Alternatively, since the cycle length is 4, we can compute 6 divided by 4, which gives a remainder of 2. Therefore, the units digit is the same as 8^2, which is 4.

All methods lead to the same conclusion.

Therefore, I can be confident that the units digit of 18^6 is 4.

**Final Answer**
The units digit of $18^6$ is \boxed{4}.

To determine the **units digit** of $ 18^6 $, we focus on the **last digit** of the result, which is determined only by the **units digit** of the base number raised to the given power.

---

### Step 1: Simplify the Problem

The number $ 18 $ has a **units digit of 8**, so we can simplify the problem to finding the **units digit of $ 8^6 $**.

This is because the **units digit of a power** depends only on the **units digit of the base**. For example:

- $ 18^2 = 324 $ → units digit is 4
- $ 8^2 = 64 $ → units digit is 4

So, we can focus on computing the units digit of $ 8^6 $.

---

### Step 2: Compute Powers of 8 and Observe the Pattern

Let’s compute the units digits of successive powers of 8:

- $ 8^1 = 8 $ → units digit = **8**
- $ 8^2 = 64 $ → units digit = **4**
- $ 8^3 = 512 $ → units digit = **2**
- $ 8^4 = 4096 $ → units digit = **6**
- $ 8^5 = 32768 $ → units digit = **8**
- $ 8^6 = 262144 $ → units digit = **4**

We observe a **repeating cycle** of units digits:
**8 → 4 → 2 → 6 → 8 → 4 → ...**

This cycle has a **length of 4**.

---

### Step 3: Use the Pattern to Find the Units Digit of $ 8^6 $

Since the cycle repeats every 4 exponents, we can find the position of the exponent 6 in the cycle by computing:

$$
6 \mod 4 = 2
$$

This means the units digit of $ 8^6 $ is the **same as the units digit of $ 8^2 $**, which is **4**.

---

### Final Answer

$$
\boxed{4}
$$

Solution: The units digit of $18^6$ is the same as in $8^6$.  There are several ways we could go about finding that units digit, but notice that $8^6 = 2^{18}$.  It's easy to find the pattern of units digits for powers of 2:  \begin{align*} 2^1 &= 2 \\ 2^2 &= 4 \\ 2^3 &= 8 \\ 2^4 &= 16 \\ 2^5 &= 32 \end{align*}Using this pattern, the units digit is found to be $\boxed{4}$.


Explanation:
Okay, so I need to figure out what kind of graph the equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$ represents. Let me start by recalling the standard forms of conic sections.

First, I remember that conic sections can be circles, ellipses, parabolas, hyperbolas, or even lines, points, or empty sets. The general idea is that the equation can be rewritten in a standard form to identify which one it is.

Looking at the given equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Hmm, this looks similar to the standard form of a circle or an ellipse. Let me recall the standard forms.

A circle has the equation $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. An ellipse has the equation $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes.

But in the given equation, there's a $\frac{x}{2}$ term. Let me try to manipulate the equation to see if it can be written in a standard form.

First, let me rewrite the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

I notice that $\frac{x}{2} - 3$ can be written as $\frac{1}{2}(x - 6)$. Let me check that:

$$
\frac{1}{2}(x - 6) = \frac{x}{2} - \frac{6}{2} = \frac{x}{2} - 3
$$

Yes, that's correct. So substituting back into the equation:

$$
\left(\frac{1}{2}(x - 6)\right)^2 + y^2 = 10
$$

Now, squaring the $\frac{1}{2}$ term:

$$
\frac{1}{4}(x - 6)^2 + y^2 = 10
$$

Okay, so now the equation is $\frac{(x - 6)^2}{4} + y^2 = 10$. Wait, if I want to write this in the standard form of an ellipse, I need to have 1 on the right-hand side. Let me divide both sides by 10 to do that:

$$
\frac{(x - 6)^2}{4 \times 10} + \frac{y^2}{10} = 1
$$

Which simplifies to:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

Now, comparing this with the standard form of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes.

In this case, the center of the ellipse is at $(h, k) = (6, 0)$. The denominators under the $x$ term is 40 and under the $y$ term is 10. Since 40 is larger than 10, the major axis is along the x-axis.

But wait, hold on. Let me check if this is indeed an ellipse. The standard form of an ellipse requires that both denominators are positive and the equation equals 1. Since both denominators are positive and we have 1 on the right-hand side, this is indeed an ellipse.

But let me make sure that I didn't make a mistake in my algebra. Let me go back through the steps again.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Expanding $\left(\frac{x}{2} - 3\right)^2$:

First, $\frac{x}{2} - 3$ squared is $\left(\frac{x}{2}\right)^2 - 2 \times \frac{x}{2} \times 3 + 3^2 = \frac{x^2}{4} - 3x + 9$.

So substituting back into the original equation:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Simplify:

$$
\frac{x^2}{4} - 3x + y^2 + 9 - 10 = 0 \\
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Hmm, this seems different from the previous form. Wait, but earlier I had rewritten the original equation as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$. Let me check if these are equivalent.

Let me take the equation $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$ and expand it:

First, expand $(x - 6)^2$:

$$
(x - 6)^2 = x^2 - 12x + 36
$$

So substituting back:

$$
\frac{x^2 - 12x + 36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{12x}{40} + \frac{36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{3x}{10} + \frac{9}{10} + \frac{y^2}{10} = 1
$$

Multiply all terms by 40 to eliminate denominators:

$$
x^2 - 12x + 36 + 4y^2 = 40 \\
x^2 - 12x + 4y^2 + 36 - 40 = 0 \\
x^2 - 12x + 4y^2 - 4 = 0
$$

Wait, but earlier when I expanded the original equation, I had:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Which is equivalent to multiplying by 4:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

Which matches the result from the other method. So both methods are consistent. Therefore, my previous steps are correct.

So, going back, the equation can be written as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, which is an ellipse with center at (6, 0), semi-major axis length $\sqrt{40}$ along the x-axis, and semi-minor axis length $\sqrt{10}$ along the y-axis.

But wait, let me check if this is indeed an ellipse. The standard form of an ellipse requires that the denominators are positive and the equation equals 1. Since both denominators are positive and the equation equals 1, this is indeed an ellipse.

Alternatively, if I think about the original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. This is similar to the equation of a circle, but with the x-term scaled by 1/2. If it were a circle, the coefficients of the squared terms would be the same. However, here, the coefficient of the x-term is 1/4 (since $\left(\frac{x}{2}\right)^2 = \frac{x^2}{4}$) and the coefficient of the y-term is 1. Therefore, it's not a circle, but an ellipse because the coefficients are different.

Alternatively, if I think about the general form of a conic section: $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. In this case, the equation is $\frac{x^2}{4} - 3x + y^2 + 9 = 10$, which simplifies to $\frac{x^2}{4} + y^2 - 3x - 1 = 0$. Multiplying through by 4 to eliminate the fraction: $x^2 + 4y^2 - 12x - 4 = 0$.

In this form, the coefficients of $x^2$ and $y^2$ are 1 and 4, respectively. The discriminant for conic sections is $B^2 - 4AC$. Here, since there is no $xy$ term, $B = 0$, so discriminant is $0 - 4(1)(4) = -16$. Since the discriminant is negative, this is an ellipse.

Therefore, regardless of the method, the equation represents an ellipse.

But let me check if there's any possibility that it could be a circle. For a circle, the coefficients of $x^2$ and $y^2$ must be equal. In the original equation, after expanding, the coefficient of $x^2$ is 1/4 and the coefficient of $y^2$ is 1. Since they are not equal, it's not a circle.

Alternatively, if we consider the original equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, if we let $u = \frac{x}{2} - 3$, then the equation becomes $u^2 + y^2 = 10$, which is a circle in the $u$-$y$ plane with radius $\sqrt{10}$. However, since $u = \frac{x}{2} - 3$, this is a horizontal scaling and shifting of the coordinate system. Therefore, the original equation is a circle in the transformed coordinates, but in the original coordinates, it's an ellipse.

So, in conclusion, the graph is an ellipse.

But let me check if there's any other possibility. For example, could this equation represent a line or something else? If the equation were a line, it would be linear in x and y, but here we have squared terms, so that's not possible. Similarly, a hyperbola would have a minus sign between the squared terms, but here we have a plus sign. A parabola would have only one squared term, but here we have two squared terms. Therefore, it's not a parabola.

Also, since the equation is equal to 10, which is a positive number, and both squared terms are non-negative, the equation is not empty. For example, if we plug in x = 6 and y = 0, we get $\left(\frac{6}{2} - 3\right)^2 + 0^2 = (3 - 3)^2 + 0 = 0 + 0 = 0$, which is not equal to 10. Wait, that's strange. Wait, hold on. Wait, if we plug in x = 6 and y = 0, we get 0 + 0 = 0, which is not 10. But according to the equation $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, plugging in x = 6 and y = 0 gives 0 + 0 = 0, which is not 1. Wait, that seems contradictory.

Wait, hold on. Let me check again.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

If I plug in x = 6 and y = 0, then:

Left-hand side: $\left(\frac{6}{2} - 3\right)^2 + 0^2 = (3 - 3)^2 + 0 = 0 + 0 = 0$, which is not equal to 10.

But according to the standard form I derived earlier, $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, plugging in x = 6 and y = 0 gives 0 + 0 = 0, which is not equal to 1.

Wait, so there's a mistake here. How come?

Wait, earlier when I divided by 10, I think I made an error. Let me check that again.

Original equation after expanding was:

$$
\frac{(x - 6)^2}{4} + y^2 = 10
$$

Wait, no. Wait, earlier steps:

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Which is $\frac{(x - 6)^2}{4} + y^2 = 10$.

So if I want to write this in standard form, I need to divide both sides by 10:

$$
\frac{(x - 6)^2}{4 \times 10} + \frac{y^2}{10} = 1
$$

Which is $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$.

But then, if I plug in x = 6 and y = 0, I get 0 + 0 = 0, which is not 1. But according to the original equation, when x = 6 and y = 0, the left-hand side is 0, which is not equal to 10. So that's consistent.

But then, how do we know that this equation actually represents a real ellipse? Because if we have $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, then the ellipse exists as long as the denominators are positive, which they are.

But to check if there are real points on the ellipse, we can find the vertices. For example, when y = 0, the equation becomes $\frac{(x - 6)^2}{40} = 1$, so $x - 6 = \pm \sqrt{40}$, so $x = 6 \pm 2\sqrt{10}$. Similarly, when x = 6, the equation becomes $\frac{y^2}{10} = 1$, so $y = \pm \sqrt{10}$. Therefore, there are real points on the graph, so it's not empty.

Therefore, the equation is indeed an ellipse.

Wait, but let me check if I made a mistake in the initial transformation. Let me take a point that should lie on the ellipse. For example, suppose x = 6 + 2√10 and y = 0. Let's plug into the original equation:

Left-hand side: $\left(\frac{6 + 2\sqrt{10}}{2} - 3\right)^2 + 0^2 = \left(3 + \sqrt{10} - 3\right)^2 = (\sqrt{10})^2 = 10$. Which matches the right-hand side. So that point is on the graph. Similarly, if I take x = 6 and y = √10, then:

Left-hand side: $\left(\frac{6}{2} - 3\right)^2 + (\sqrt{10})^2 = (3 - 3)^2 + 10 = 0 + 10 = 10$. Correct.

Therefore, the equation does represent a real ellipse.

So, after all these checks, I can conclude that the graph is an ellipse.

But let me just make sure that there isn't a case where this could be a different conic. For example, if the equation had a negative sign, it would be a hyperbola. But here, both terms are positive, so it's an ellipse.

Another way to think about it: the equation is a sum of squares equals a positive constant. In general, equations of the form $A(x - h)^2 + B(y - k)^2 = C$ where A and B are positive and C is positive represent ellipses if A ≠ B, and circles if A = B.

In this case, A is 1/4 (from $\left(\frac{x}{2} - 3\right)^2$) and B is 1 (from $y^2$). Since 1/4 ≠ 1, it's an ellipse.

Therefore, the answer should be an ellipse.

**Final Answer**
The graph of the equation is an \boxed{ellipse}.

To determine the type of graph represented by the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

we begin by analyzing its structure and comparing it to the standard forms of conic sections.

---

### Step 1: Recognize the Form

The equation contains two squared terms, one involving $ x $ and the other involving $ y $, and the right-hand side is a constant. This suggests the possibility of a **circle** or **ellipse**, as both are conic sections with squared terms.

However, the coefficient of the $ x $-term is not 1, which is a key difference between a circle and an ellipse. A **circle** has the form:

$$
(x - h)^2 + (y - k)^2 = r^2
$$

where the coefficients of the squared terms are equal. In contrast, an **ellipse** has the form:

$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$

where the denominators are not necessarily equal.

---

### Step 2: Rewrite the Equation

We start by simplifying the given equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

Note that:

$$
\frac{x}{2} - 3 = \frac{1}{2}(x - 6)
$$

So the equation becomes:

$$
\left(\frac{1}{2}(x - 6)\right)^2 + y^2 = 10
$$

Squaring the $ \frac{1}{2} $ term:

$$
\frac{(x - 6)^2}{4} + y^2 = 10
$$

Now, to write this in standard form, divide both sides by 10:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

This is now in the standard form of an **ellipse**:

$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$

where:
- $ h = 6 $, $ k = 0 $ (center of the ellipse)
- $ a^2 = 40 $, $ b^2 = 10 $

So the semi-major axis is along the $ x $-axis (since $ a^2 > b^2 $), and the ellipse is centered at $ (6, 0) $.

---

### Step 3: Confirm the Type of Conic

To further confirm, we can use the **discriminant test** for conic sections. The general form of a conic is:

$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
$$

The discriminant is:

$$
\Delta = B^2 - 4AC
$$

In our case, the equation is:

$$
\frac{x^2}{4} + y^2 - 3x - 1 = 0
$$

Multiplying through by 4 to eliminate the fraction:

$$
x^2 + 4y^2 - 12x - 4 = 0
$$

Here, $ A = 1 $, $ B = 0 $, $ C = 4 $, so:

$$
\Delta = 0^2 - 4(1)(4) = -16
$$

Since $ \Delta < 0 $, the conic is an **ellipse**.

---

### Step 4: Eliminate Other Possibilities

- **Circle**: Requires equal coefficients for $ x^2 $ and $ y^2 $, which is not the case here.
- **Parabola**: Would have only one squared term, which is not the case.
- **Hyperbola**: Would have a minus sign between the squared terms, which is not the case.
- **Line or Point**: These would not involve squared terms.
- **Empty Set**: Not the case, as we can find real points on the graph (e.g., $ x = 6 + 2\sqrt{10}, y = 0 $).

---

### Final Conclusion

The given equation represents a **non-degenerate ellipse** with center at $ (6, 0) $, semi-major axis $ \sqrt{40} $, and semi-minor axis $ \sqrt{10} $.

$$
\boxed{ellipse}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what kind of graph the equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$ represents. Let me start by recalling the standard forms of conic sections.

First, I remember that conic sections can be circles, ellipses, parabolas, hyperbolas, or even lines, points, or empty sets. The general idea is that the equation can be rewritten in a standard form to identify which one it is.

Looking at the given equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. Hmm, this looks similar to the standard form of a circle or an ellipse. Let me recall the standard forms.

A circle has the equation $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. An ellipse has the equation $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes.

But in the given equation, there's a $\frac{x}{2}$ term. Let me try to manipulate the equation to see if it can be written in a standard form.

First, let me rewrite the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

I notice that $\frac{x}{2} - 3$ can be written as $\frac{1}{2}(x - 6)$. Let me check that:

$$
\frac{1}{2}(x - 6) = \frac{x}{2} - \frac{6}{2} = \frac{x}{2} - 3
$$

Yes, that's correct. So substituting back into the equation:

$$
\left(\frac{1}{2}(x - 6)\right)^2 + y^2 = 10
$$

Now, squaring the $\frac{1}{2}$ term:

$$
\frac{1}{4}(x - 6)^2 + y^2 = 10
$$

Okay, so now the equation is $\frac{(x - 6)^2}{4} + y^2 = 10$. Wait, if I want to write this in the standard form of an ellipse, I need to have 1 on the right-hand side. Let me divide both sides by 10 to do that:

$$
\frac{(x - 6)^2}{4 \times 10} + \frac{y^2}{10} = 1
$$

Which simplifies to:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

Now, comparing this with the standard form of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes.

In this case, the center of the ellipse is at $(h, k) = (6, 0)$. The denominators under the $x$ term is 40 and under the $y$ term is 10. Since 40 is larger than 10, the major axis is along the x-axis.

But wait, hold on. Let me check if this is indeed an ellipse. The standard form of an ellipse requires that both denominators are positive and the equation equals 1. Since both denominators are positive and we have 1 on the right-hand side, this is indeed an ellipse.

But let me make sure that I didn't make a mistake in my algebra. Let me go back through the steps again.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Expanding $\left(\frac{x}{2} - 3\right)^2$:

First, $\frac{x}{2} - 3$ squared is $\left(\frac{x}{2}\right)^2 - 2 \times \frac{x}{2} \times 3 + 3^2 = \frac{x^2}{4} - 3x + 9$.

So substituting back into the original equation:

$$
\frac{x^2}{4} - 3x + 9 + y^2 = 10
$$

Simplify:

$$
\frac{x^2}{4} - 3x + y^2 + 9 - 10 = 0 \\
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Hmm, this seems different from the previous form. Wait, but earlier I had rewritten the original equation as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$. Let me check if these are equivalent.

Let me take the equation $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$ and expand it:

First, expand $(x - 6)^2$:

$$
(x - 6)^2 = x^2 - 12x + 36
$$

So substituting back:

$$
\frac{x^2 - 12x + 36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{12x}{40} + \frac{36}{40} + \frac{y^2}{10} = 1 \\
\frac{x^2}{40} - \frac{3x}{10} + \frac{9}{10} + \frac{y^2}{10} = 1
$$

Multiply all terms by 40 to eliminate denominators:

$$
x^2 - 12x + 36 + 4y^2 = 40 \\
x^2 - 12x + 4y^2 + 36 - 40 = 0 \\
x^2 - 12x + 4y^2 - 4 = 0
$$

Wait, but earlier when I expanded the original equation, I had:

$$
\frac{x^2}{4} - 3x + y^2 - 1 = 0
$$

Which is equivalent to multiplying by 4:

$$
x^2 - 12x + 4y^2 - 4 = 0
$$

Which matches the result from the other method. So both methods are consistent. Therefore, my previous steps are correct.

So, going back, the equation can be written as $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, which is an ellipse with center at (6, 0), semi-major axis length $\sqrt{40}$ along the x-axis, and semi-minor axis length $\sqrt{10}$ along the y-axis.

But wait, let me check if this is indeed an ellipse. The standard form of an ellipse requires that the denominators are positive and the equation equals 1. Since both denominators are positive and the equation equals 1, this is indeed an ellipse.

Alternatively, if I think about the original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$. This is similar to the equation of a circle, but with the x-term scaled by 1/2. If it were a circle, the coefficients of the squared terms would be the same. However, here, the coefficient of the x-term is 1/4 (since $\left(\frac{x}{2}\right)^2 = \frac{x^2}{4}$) and the coefficient of the y-term is 1. Therefore, it's not a circle, but an ellipse because the coefficients are different.

Alternatively, if I think about the general form of a conic section: $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. In this case, the equation is $\frac{x^2}{4} - 3x + y^2 + 9 = 10$, which simplifies to $\frac{x^2}{4} + y^2 - 3x - 1 = 0$. Multiplying through by 4 to eliminate the fraction: $x^2 + 4y^2 - 12x - 4 = 0$.

In this form, the coefficients of $x^2$ and $y^2$ are 1 and 4, respectively. The discriminant for conic sections is $B^2 - 4AC$. Here, since there is no $xy$ term, $B = 0$, so discriminant is $0 - 4(1)(4) = -16$. Since the discriminant is negative, this is an ellipse.

Therefore, regardless of the method, the equation represents an ellipse.

But let me check if there's any possibility that it could be a circle. For a circle, the coefficients of $x^2$ and $y^2$ must be equal. In the original equation, after expanding, the coefficient of $x^2$ is 1/4 and the coefficient of $y^2$ is 1. Since they are not equal, it's not a circle.

Alternatively, if we consider the original equation $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$, if we let $u = \frac{x}{2} - 3$, then the equation becomes $u^2 + y^2 = 10$, which is a circle in the $u$-$y$ plane with radius $\sqrt{10}$. However, since $u = \frac{x}{2} - 3$, this is a horizontal scaling and shifting of the coordinate system. Therefore, the original equation is a circle in the transformed coordinates, but in the original coordinates, it's an ellipse.

So, in conclusion, the graph is an ellipse.

But let me check if there's any other possibility. For example, could this equation represent a line or something else? If the equation were a line, it would be linear in x and y, but here we have squared terms, so that's not possible. Similarly, a hyperbola would have a minus sign between the squared terms, but here we have a plus sign. A parabola would have only one squared term, but here we have two squared terms. Therefore, it's not a parabola.

Also, since the equation is equal to 10, which is a positive number, and both squared terms are non-negative, the equation is not empty. For example, if we plug in x = 6 and y = 0, we get $\left(\frac{6}{2} - 3\right)^2 + 0^2 = (3 - 3)^2 + 0 = 0 + 0 = 0$, which is not equal to 10. Wait, that's strange. Wait, hold on. Wait, if we plug in x = 6 and y = 0, we get 0 + 0 = 0, which is not 10. But according to the equation $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, plugging in x = 6 and y = 0 gives 0 + 0 = 0, which is not 1. Wait, that seems contradictory.

Wait, hold on. Let me check again.

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

If I plug in x = 6 and y = 0, then:

Left-hand side: $\left(\frac{6}{2} - 3\right)^2 + 0^2 = (3 - 3)^2 + 0 = 0 + 0 = 0$, which is not equal to 10.

But according to the standard form I derived earlier, $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, plugging in x = 6 and y = 0 gives 0 + 0 = 0, which is not equal to 1.

Wait, so there's a mistake here. How come?

Wait, earlier when I divided by 10, I think I made an error. Let me check that again.

Original equation after expanding was:

$$
\frac{(x - 6)^2}{4} + y^2 = 10
$$

Wait, no. Wait, earlier steps:

Original equation: $\left(\frac{x}{2} - 3\right)^2 + y^2 = 10$.

Which is $\frac{(x - 6)^2}{4} + y^2 = 10$.

So if I want to write this in standard form, I need to divide both sides by 10:

$$
\frac{(x - 6)^2}{4 \times 10} + \frac{y^2}{10} = 1
$$

Which is $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$.

But then, if I plug in x = 6 and y = 0, I get 0 + 0 = 0, which is not 1. But according to the original equation, when x = 6 and y = 0, the left-hand side is 0, which is not equal to 10. So that's consistent.

But then, how do we know that this equation actually represents a real ellipse? Because if we have $\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1$, then the ellipse exists as long as the denominators are positive, which they are.

But to check if there are real points on the ellipse, we can find the vertices. For example, when y = 0, the equation becomes $\frac{(x - 6)^2}{40} = 1$, so $x - 6 = \pm \sqrt{40}$, so $x = 6 \pm 2\sqrt{10}$. Similarly, when x = 6, the equation becomes $\frac{y^2}{10} = 1$, so $y = \pm \sqrt{10}$. Therefore, there are real points on the graph, so it's not empty.

Therefore, the equation is indeed an ellipse.

Wait, but let me check if I made a mistake in the initial transformation. Let me take a point that should lie on the ellipse. For example, suppose x = 6 + 2√10 and y = 0. Let's plug into the original equation:

Left-hand side: $\left(\frac{6 + 2\sqrt{10}}{2} - 3\right)^2 + 0^2 = \left(3 + \sqrt{10} - 3\right)^2 = (\sqrt{10})^2 = 10$. Which matches the right-hand side. So that point is on the graph. Similarly, if I take x = 6 and y = √10, then:

Left-hand side: $\left(\frac{6}{2} - 3\right)^2 + (\sqrt{10})^2 = (3 - 3)^2 + 10 = 0 + 10 = 10$. Correct.

Therefore, the equation does represent a real ellipse.

So, after all these checks, I can conclude that the graph is an ellipse.

But let me just make sure that there isn't a case where this could be a different conic. For example, if the equation had a negative sign, it would be a hyperbola. But here, both terms are positive, so it's an ellipse.

Another way to think about it: the equation is a sum of squares equals a positive constant. In general, equations of the form $A(x - h)^2 + B(y - k)^2 = C$ where A and B are positive and C is positive represent ellipses if A ≠ B, and circles if A = B.

In this case, A is 1/4 (from $\left(\frac{x}{2} - 3\right)^2$) and B is 1 (from $y^2$). Since 1/4 ≠ 1, it's an ellipse.

Therefore, the answer should be an ellipse.

**Final Answer**
The graph of the equation is an \boxed{ellipse}.

To determine the type of graph represented by the equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

we begin by analyzing its structure and comparing it to the standard forms of conic sections.

---

### Step 1: Recognize the Form

The equation contains two squared terms, one involving $ x $ and the other involving $ y $, and the right-hand side is a constant. This suggests the possibility of a **circle** or **ellipse**, as both are conic sections with squared terms.

However, the coefficient of the $ x $-term is not 1, which is a key difference between a circle and an ellipse. A **circle** has the form:

$$
(x - h)^2 + (y - k)^2 = r^2
$$

where the coefficients of the squared terms are equal. In contrast, an **ellipse** has the form:

$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$

where the denominators are not necessarily equal.

---

### Step 2: Rewrite the Equation

We start by simplifying the given equation:

$$
\left(\frac{x}{2} - 3\right)^2 + y^2 = 10
$$

Note that:

$$
\frac{x}{2} - 3 = \frac{1}{2}(x - 6)
$$

So the equation becomes:

$$
\left(\frac{1}{2}(x - 6)\right)^2 + y^2 = 10
$$

Squaring the $ \frac{1}{2} $ term:

$$
\frac{(x - 6)^2}{4} + y^2 = 10
$$

Now, to write this in standard form, divide both sides by 10:

$$
\frac{(x - 6)^2}{40} + \frac{y^2}{10} = 1
$$

This is now in the standard form of an **ellipse**:

$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$

where:
- $ h = 6 $, $ k = 0 $ (center of the ellipse)
- $ a^2 = 40 $, $ b^2 = 10 $

So the semi-major axis is along the $ x $-axis (since $ a^2 > b^2 $), and the ellipse is centered at $ (6, 0) $.

---

### Step 3: Confirm the Type of Conic

To further confirm, we can use the **discriminant test** for conic sections. The general form of a conic is:

$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
$$

The discriminant is:

$$
\Delta = B^2 - 4AC
$$

In our case, the equation is:

$$
\frac{x^2}{4} + y^2 - 3x - 1 = 0
$$

Multiplying through by 4 to eliminate the fraction:

$$
x^2 + 4y^2 - 12x - 4 = 0
$$

Here, $ A = 1 $, $ B = 0 $, $ C = 4 $, so:

$$
\Delta = 0^2 - 4(1)(4) = -16
$$

Since $ \Delta < 0 $, the conic is an **ellipse**.

---

### Step 4: Eliminate Other Possibilities

- **Circle**: Requires equal coefficients for $ x^2 $ and $ y^2 $, which is not the case here.
- **Parabola**: Would have only one squared term, which is not the case.
- **Hyperbola**: Would have a minus sign between the squared terms, which is not the case.
- **Line or Point**: These would not involve squared terms.
- **Empty Set**: Not the case, as we can find real points on the graph (e.g., $ x = 6 + 2\sqrt{10}, y = 0 $).

---

### Final Conclusion

The given equation represents a **non-degenerate ellipse** with center at $ (6, 0) $, semi-major axis $ \sqrt{40} $, and semi-minor axis $ \sqrt{10} $.

$$
\boxed{ellipse}
$$

Solution: This looks like the equation of a circle, but we have replaced $x$ with $\frac x2$. So, we suspect this equation defines an $\boxed{\text{ellipse}}$. To verify this we write \[\left(\frac x2 - 3\right)^2 = \frac 14 \left( x - 6\right)^2,\]and we see that the equation \[ \frac{\left(x - 6 \right)^2}{4} + y^2 = 10 \]is the equation of an ellipse.


Explanation:
Okay, so I need to solve this equation: Three plus the reciprocal of a number equals 7 divided by that number. Let me try to write that down first. Let me call the number x. So, translating the words into math, it would be 3 plus 1 over x equals 7 over x. Let me check that again. "Three plus the reciprocal of a number" – reciprocal of x is 1/x, so 3 + 1/x. Then "equals 7 divided by that number," which is 7/x. So the equation is 3 + (1/x) = 7/x.

Alright, so now I need to solve for x. Let me write that equation again:

3 + (1/x) = 7/x

Hmm, okay. Let me think about how to solve this. Since there are fractions here, maybe I can eliminate the denominators by multiplying both sides by x. That might make it easier. Let me try that. If I multiply every term by x, then:

x * 3 + x * (1/x) = x * (7/x)

Simplifying each term:

First term: 3x

Second term: x*(1/x) is 1

Third term: x*(7/x) is 7

So the equation becomes:

3x + 1 = 7

Okay, that seems simpler. Now, solving for x. Let me subtract 1 from both sides:

3x + 1 - 1 = 7 - 1

Which simplifies to:

3x = 6

Then divide both sides by 3:

x = 6 / 3

So x = 2.

Wait, let me check if that's correct. Let me plug x = 2 back into the original equation to verify.

Original equation: 3 + (1/x) = 7/x

Left side: 3 + (1/2) = 3.5 or 7/2

Right side: 7 / 2 = 3.5

Yes, both sides are equal. So x = 2 is the solution.

But wait, let me think again. Did I do everything right? Let me go through the steps again. Starting with 3 + 1/x = 7/x. Multiply both sides by x: 3x + 1 = 7. Then 3x = 6, so x = 2. Yep, that seems correct.

But maybe I should consider if there are any restrictions on x? Since there's a reciprocal, x can't be zero. So x ≠ 0. But in this case, our solution is x = 2, which is fine. So no issues there.

Alternatively, maybe I can solve it another way to confirm. Let me try rearranging the original equation.

Starting with 3 + 1/x = 7/x. Let me subtract 1/x from both sides:

3 = 7/x - 1/x

Which simplifies to 3 = (7 - 1)/x = 6/x

So 3 = 6/x. Then solving for x, multiply both sides by x:

3x = 6 => x = 2. Same result.

Okay, so that confirms it again. So x = 2 is the solution.

Another way: Let me think. Maybe combining the terms on the left side. The original equation is 3 + 1/x = 7/x. Let me subtract 7/x from both sides:

3 + 1/x - 7/x = 0

Which simplifies to 3 - 6/x = 0

Then, 3 = 6/x, same as before. So again, x = 2.

So all methods lead to x = 2. Therefore, I think that's the correct answer.

But just to be thorough, let me check if there's any other possible solution. For example, sometimes when you multiply both sides by a variable, you might introduce extraneous solutions, but in this case, since we only had one solution and it checked out, I don't think there are others.

Alternatively, suppose I didn't multiply by x, but instead tried to combine the terms. Let me try that.

Original equation: 3 + 1/x = 7/x. Let me subtract 3 from both sides:

1/x = 7/x - 3

Hmm, not sure if that helps. Let me write 7/x - 3 as 7/x - 3x/x = (7 - 3x)/x. Therefore:

1/x = (7 - 3x)/x

Multiply both sides by x:

1 = 7 - 3x

Then, subtract 7 from both sides:

1 - 7 = -3x => -6 = -3x => x = (-6)/(-3) = 2. Same answer again.

So regardless of the method, I get x = 2.

Therefore, I can be confident that the number is 2.

Just to recap, the key steps were translating the word problem into an equation, then solving the equation by eliminating denominators through multiplication, simplifying, and solving the resulting linear equation. Then verifying the solution by plugging it back into the original equation. All steps led to the same answer, so I think that's solid.

I don't see any mistakes in my reasoning, so I think the answer is correct.

**Final Answer**
The number is \boxed{2}.

To solve the problem:

> "Three plus the reciprocal of a number equals 7 divided by that number,"

we begin by translating the words into a mathematical equation. Let the unknown number be $ x $. The reciprocal of $ x $ is $ \frac{1}{x} $, and the equation becomes:

$$
3 + \frac{1}{x} = \frac{7}{x}
$$

---

### Step 1: Eliminate the Denominators

To simplify the equation, we can eliminate the denominators by multiplying both sides of the equation by $ x $, assuming $ x \neq 0 $ (since division by zero is undefined):

$$
x \cdot \left(3 + \frac{1}{x}\right) = x \cdot \frac{7}{x}
$$

Distribute $ x $ on the left-hand side:

$$
3x + 1 = 7
$$

---

### Step 2: Solve the Simplified Equation

Now solve for $ x $:

$$
3x + 1 = 7
$$

Subtract 1 from both sides:

$$
3x = 6
$$

Divide both sides by 3:

$$
x = 2
$$

---

### Step 3: Verify the Solution

Plug $ x = 2 $ back into the original equation to confirm:

- Left-hand side: $ 3 + \frac{1}{2} = 3.5 $
- Right-hand side: $ \frac{7}{2} = 3.5 $

Both sides are equal, so the solution is valid.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: Three plus the reciprocal of a number equals 7 divided by that number. Let me try to write that down first. Let me call the number x. So, translating the words into math, it would be 3 plus 1 over x equals 7 over x. Let me check that again. "Three plus the reciprocal of a number" – reciprocal of x is 1/x, so 3 + 1/x. Then "equals 7 divided by that number," which is 7/x. So the equation is 3 + (1/x) = 7/x.

Alright, so now I need to solve for x. Let me write that equation again:

3 + (1/x) = 7/x

Hmm, okay. Let me think about how to solve this. Since there are fractions here, maybe I can eliminate the denominators by multiplying both sides by x. That might make it easier. Let me try that. If I multiply every term by x, then:

x * 3 + x * (1/x) = x * (7/x)

Simplifying each term:

First term: 3x

Second term: x*(1/x) is 1

Third term: x*(7/x) is 7

So the equation becomes:

3x + 1 = 7

Okay, that seems simpler. Now, solving for x. Let me subtract 1 from both sides:

3x + 1 - 1 = 7 - 1

Which simplifies to:

3x = 6

Then divide both sides by 3:

x = 6 / 3

So x = 2.

Wait, let me check if that's correct. Let me plug x = 2 back into the original equation to verify.

Original equation: 3 + (1/x) = 7/x

Left side: 3 + (1/2) = 3.5 or 7/2

Right side: 7 / 2 = 3.5

Yes, both sides are equal. So x = 2 is the solution.

But wait, let me think again. Did I do everything right? Let me go through the steps again. Starting with 3 + 1/x = 7/x. Multiply both sides by x: 3x + 1 = 7. Then 3x = 6, so x = 2. Yep, that seems correct.

But maybe I should consider if there are any restrictions on x? Since there's a reciprocal, x can't be zero. So x ≠ 0. But in this case, our solution is x = 2, which is fine. So no issues there.

Alternatively, maybe I can solve it another way to confirm. Let me try rearranging the original equation.

Starting with 3 + 1/x = 7/x. Let me subtract 1/x from both sides:

3 = 7/x - 1/x

Which simplifies to 3 = (7 - 1)/x = 6/x

So 3 = 6/x. Then solving for x, multiply both sides by x:

3x = 6 => x = 2. Same result.

Okay, so that confirms it again. So x = 2 is the solution.

Another way: Let me think. Maybe combining the terms on the left side. The original equation is 3 + 1/x = 7/x. Let me subtract 7/x from both sides:

3 + 1/x - 7/x = 0

Which simplifies to 3 - 6/x = 0

Then, 3 = 6/x, same as before. So again, x = 2.

So all methods lead to x = 2. Therefore, I think that's the correct answer.

But just to be thorough, let me check if there's any other possible solution. For example, sometimes when you multiply both sides by a variable, you might introduce extraneous solutions, but in this case, since we only had one solution and it checked out, I don't think there are others.

Alternatively, suppose I didn't multiply by x, but instead tried to combine the terms. Let me try that.

Original equation: 3 + 1/x = 7/x. Let me subtract 3 from both sides:

1/x = 7/x - 3

Hmm, not sure if that helps. Let me write 7/x - 3 as 7/x - 3x/x = (7 - 3x)/x. Therefore:

1/x = (7 - 3x)/x

Multiply both sides by x:

1 = 7 - 3x

Then, subtract 7 from both sides:

1 - 7 = -3x => -6 = -3x => x = (-6)/(-3) = 2. Same answer again.

So regardless of the method, I get x = 2.

Therefore, I can be confident that the number is 2.

Just to recap, the key steps were translating the word problem into an equation, then solving the equation by eliminating denominators through multiplication, simplifying, and solving the resulting linear equation. Then verifying the solution by plugging it back into the original equation. All steps led to the same answer, so I think that's solid.

I don't see any mistakes in my reasoning, so I think the answer is correct.

**Final Answer**
The number is \boxed{2}.

To solve the problem:

> "Three plus the reciprocal of a number equals 7 divided by that number,"

we begin by translating the words into a mathematical equation. Let the unknown number be $ x $. The reciprocal of $ x $ is $ \frac{1}{x} $, and the equation becomes:

$$
3 + \frac{1}{x} = \frac{7}{x}
$$

---

### Step 1: Eliminate the Denominators

To simplify the equation, we can eliminate the denominators by multiplying both sides of the equation by $ x $, assuming $ x \neq 0 $ (since division by zero is undefined):

$$
x \cdot \left(3 + \frac{1}{x}\right) = x \cdot \frac{7}{x}
$$

Distribute $ x $ on the left-hand side:

$$
3x + 1 = 7
$$

---

### Step 2: Solve the Simplified Equation

Now solve for $ x $:

$$
3x + 1 = 7
$$

Subtract 1 from both sides:

$$
3x = 6
$$

Divide both sides by 3:

$$
x = 2
$$

---

### Step 3: Verify the Solution

Plug $ x = 2 $ back into the original equation to confirm:

- Left-hand side: $ 3 + \frac{1}{2} = 3.5 $
- Right-hand side: $ \frac{7}{2} = 3.5 $

Both sides are equal, so the solution is valid.

---

### Final Answer

$$
\boxed{2}
$$

Solution: Let $x$ be the number.  Converting the words in the problem into an equation gives us $3+\dfrac{1}{x} = \dfrac{7}{x}$.  Subtracting $\dfrac{1}{x}$ from both sides gives $3 = \dfrac{6}{x}$. Multiplying both sides of this equation by $x$ gives $3x =6$, and dividing both sides of this equation by 3 gives $x = \boxed{2}$.


Explanation:
Okay, so I need to find the greatest possible perimeter of a triangle where two sides are each 8 units long, and the third side has a whole number length. Let me think about how to approach this.

First, I remember that in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This is called the triangle inequality theorem. So, if two sides are 8 units each, let me denote the third side as 'x'.

So, the triangle has sides 8, 8, and x. Now, applying the triangle inequality theorem, I need to make sure that the sum of any two sides is greater than the third side. Let me write down the inequalities:

1. 8 + 8 > x
2. 8 + x > 8
3. 8 + x > 8

Wait, the second and third inequalities seem similar. Let me check that again.

Actually, the three inequalities should be:

1. 8 + 8 > x
2. 8 + x > 8
3. 8 + x > 8

But since the two sides are both 8, the second and third inequalities are the same. So, simplifying:

First inequality: 16 > x, which means x < 16.

Second and third inequalities: 8 + x > 8. Subtracting 8 from both sides gives x > 0.

But since x is a length, it must be positive anyway. However, since we are told that x is a whole number, x must be at least 1. But wait, maybe there's more to it? Let me think again.

Wait, actually, the triangle inequality says that the sum of any two sides must be greater than the third side. So, for sides 8, 8, and x:

- 8 + 8 > x
- 8 + x > 8
- 8 + x > 8

So, the first inequality gives x < 16, and the other two inequalities give x > 0. But since x is a length, it must be positive. However, in reality, for the triangle to exist, x must satisfy both inequalities. But since x has to be a positive number, but also less than 16. However, since we are looking for the greatest possible perimeter, we need the maximum possible value of x.

But wait, if x can be as large as just under 16, but since x has to be a whole number, the maximum possible value of x is 15. Because if x is 16, then 8 + 8 is equal to 16, which would make the triangle degenerate (i.e., a straight line), not a valid triangle. Therefore, x must be less than 16, so the maximum whole number value for x is 15.

Therefore, the perimeter would be 8 + 8 + 15 = 31 units.

But let me check if that's correct. Let me verify with the triangle inequality. If x is 15, then:

- 8 + 8 > 15? 16 > 15, which is true.
- 8 + 15 > 8? 23 > 8, which is true.
- 8 + 15 > 8? Same as above, true.

So, yes, 15 is a valid length for the third side. Therefore, the maximum perimeter would be 8 + 8 + 15 = 31.

But wait, let me make sure there isn't a larger possible perimeter. Suppose x is 16, but as mentioned before, that would make the triangle degenerate. So, 16 is not allowed. Therefore, 15 is the maximum.

But let me think again. Is there any other constraint? For example, sometimes people might think about the third side being less than the sum of the other two sides, but in this case, since two sides are equal, maybe there's something else?

Wait, no. The triangle inequality is straightforward here. The third side must be less than the sum of the other two sides, which is 16, and greater than the difference of the other two sides. Wait, hold on, is there another inequality?

Wait, actually, another way to think about the triangle inequality is that the third side must be greater than the difference of the other two sides. Since the two sides are both 8, the difference is 8 - 8 = 0. Therefore, the third side must be greater than 0. Which again gives x > 0.

So, combining both inequalities, x must be between 0 and 16, but since x is a whole number, it can be from 1 to 15. Therefore, the maximum x is 15, so the maximum perimeter is 8 + 8 + 15 = 31.

But let me check with an example. If x is 15, then the sides are 8, 8, 15. Let me check if that's a valid triangle.

Yes, because 8 + 8 > 15 (16 > 15), which is true. The other inequalities are trivial.

Alternatively, if x was 16, then 8 + 8 = 16, which is not greater, so it's invalid.

Therefore, 15 is the maximum possible value for x.

But wait, let me think again. Suppose someone says that the third side must be less than the sum of the other two sides, but also greater than the difference. Wait, the difference here is 8 - 8 = 0, so x must be greater than 0. So, x can be 1 to 15.

But maybe there's a mistake here? Let me check with another example. Suppose the two sides are 8 and 8, and the third side is 1. Then, 8 + 1 > 8? 9 > 8, which is true. Similarly, 8 + 8 > 1, which is true. So, that works.

So, the third side can be as small as 1, and as large as 15. Therefore, the maximum perimeter is 8 + 8 + 15 = 31.

But let me check if there's a different interpretation of the problem. The problem says "two sides of a triangle are each 8 units long." So, that means two sides are 8, and the third is something else. So, the triangle is isoceles with two sides of 8.

Therefore, the answer should be 31.

But let me check again. Maybe I made a mistake in the triangle inequality. Let me recall the exact statement.

The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

So, for sides a, b, c:

a + b > c

a + c > b

b + c > a

So, in our case, sides are 8, 8, x.

So:

First inequality: 8 + 8 > x => 16 > x

Second inequality: 8 + x > 8 => x > 0

Third inequality: 8 + x > 8 => same as second, x > 0

Therefore, x must be greater than 0 and less than 16. Since x is a whole number, the maximum value is 15.

Therefore, perimeter is 8 + 8 + 15 = 31.

Therefore, the answer is 31.

But let me check if there's a possibility that the third side could be longer? For example, if the triangle was not isoceles? Wait, no, the problem says two sides are each 8 units long. So, the third side is different.

Alternatively, maybe the problem is in another way? For example, if the third side is also 8, then the triangle is equilateral, but then the perimeter would be 24. But since we are looking for the greatest possible perimeter, 31 is bigger than 24, so 31 is better.

Therefore, I think 31 is the correct answer.

But let me check once more. Suppose x is 15. Then, the sides are 8, 8, 15. Let me check if that's a valid triangle.

Yes, because 8 + 8 > 15 (16 > 15). The other inequalities are 8 + 15 > 8 (23 > 8) and 8 + 15 > 8 (same). So, all inequalities are satisfied.

Alternatively, if x was 16, then 8 + 8 = 16, which is not greater, so it's invalid.

Therefore, 15 is the maximum.

Therefore, the greatest possible perimeter is 8 + 8 + 15 = 31.

I think that's solid. So, the answer is 31.

**Final Answer**
The greatest possible perimeter is \boxed{31}.

To determine the **greatest possible perimeter** of a triangle where **two sides are each 8 units long**, and the **third side has a whole number length**, we must apply the **triangle inequality theorem**.

---

### Step 1: Understand the Triangle Inequality

The triangle inequality theorem states that for any triangle with sides $ a $, $ b $, and $ c $, the following must be true:

$$
a + b > c \\
a + c > b \\
b + c > a
$$

In this case, the triangle has sides of lengths $ 8 $, $ 8 $, and $ x $, where $ x $ is the unknown third side (a whole number). Applying the triangle inequality:

1. $ 8 + 8 > x \Rightarrow 16 > x $
2. $ 8 + x > 8 \Rightarrow x > 0 $
3. $ 8 + x > 8 \Rightarrow x > 0 $

So, combining these, we find that:

$$
0 < x < 16
$$

Since $ x $ must be a **whole number**, the possible values for $ x $ are:

$$
x = 1, 2, 3, \dots, 15
$$

---

### Step 2: Maximize the Perimeter

The perimeter $ P $ of the triangle is:

$$
P = 8 + 8 + x = 16 + x
$$

To **maximize** the perimeter, we need to **maximize** $ x $, subject to the constraint $ x < 16 $. The largest whole number less than 16 is:

$$
x = 15
$$

---

### Step 3: Verify the Triangle Inequality for $ x = 15 $

Check if the triangle with sides $ 8, 8, 15 $ is valid:

- $ 8 + 8 = 16 > 15 $ ✅
- $ 8 + 15 = 23 > 8 $ ✅
- $ 8 + 15 = 23 > 8 $ ✅

All triangle inequalities are satisfied, so this is a valid triangle.

---

### Step 4: Compute the Perimeter

$$
P = 8 + 8 + 15 = 31
$$

---

### Final Answer

$$
\boxed{31}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible perimeter of a triangle where two sides are each 8 units long, and the third side has a whole number length. Let me think about how to approach this.

First, I remember that in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This is called the triangle inequality theorem. So, if two sides are 8 units each, let me denote the third side as 'x'.

So, the triangle has sides 8, 8, and x. Now, applying the triangle inequality theorem, I need to make sure that the sum of any two sides is greater than the third side. Let me write down the inequalities:

1. 8 + 8 > x
2. 8 + x > 8
3. 8 + x > 8

Wait, the second and third inequalities seem similar. Let me check that again.

Actually, the three inequalities should be:

1. 8 + 8 > x
2. 8 + x > 8
3. 8 + x > 8

But since the two sides are both 8, the second and third inequalities are the same. So, simplifying:

First inequality: 16 > x, which means x < 16.

Second and third inequalities: 8 + x > 8. Subtracting 8 from both sides gives x > 0.

But since x is a length, it must be positive anyway. However, since we are told that x is a whole number, x must be at least 1. But wait, maybe there's more to it? Let me think again.

Wait, actually, the triangle inequality says that the sum of any two sides must be greater than the third side. So, for sides 8, 8, and x:

- 8 + 8 > x
- 8 + x > 8
- 8 + x > 8

So, the first inequality gives x < 16, and the other two inequalities give x > 0. But since x is a length, it must be positive. However, in reality, for the triangle to exist, x must satisfy both inequalities. But since x has to be a positive number, but also less than 16. However, since we are looking for the greatest possible perimeter, we need the maximum possible value of x.

But wait, if x can be as large as just under 16, but since x has to be a whole number, the maximum possible value of x is 15. Because if x is 16, then 8 + 8 is equal to 16, which would make the triangle degenerate (i.e., a straight line), not a valid triangle. Therefore, x must be less than 16, so the maximum whole number value for x is 15.

Therefore, the perimeter would be 8 + 8 + 15 = 31 units.

But let me check if that's correct. Let me verify with the triangle inequality. If x is 15, then:

- 8 + 8 > 15? 16 > 15, which is true.
- 8 + 15 > 8? 23 > 8, which is true.
- 8 + 15 > 8? Same as above, true.

So, yes, 15 is a valid length for the third side. Therefore, the maximum perimeter would be 8 + 8 + 15 = 31.

But wait, let me make sure there isn't a larger possible perimeter. Suppose x is 16, but as mentioned before, that would make the triangle degenerate. So, 16 is not allowed. Therefore, 15 is the maximum.

But let me think again. Is there any other constraint? For example, sometimes people might think about the third side being less than the sum of the other two sides, but in this case, since two sides are equal, maybe there's something else?

Wait, no. The triangle inequality is straightforward here. The third side must be less than the sum of the other two sides, which is 16, and greater than the difference of the other two sides. Wait, hold on, is there another inequality?

Wait, actually, another way to think about the triangle inequality is that the third side must be greater than the difference of the other two sides. Since the two sides are both 8, the difference is 8 - 8 = 0. Therefore, the third side must be greater than 0. Which again gives x > 0.

So, combining both inequalities, x must be between 0 and 16, but since x is a whole number, it can be from 1 to 15. Therefore, the maximum x is 15, so the maximum perimeter is 8 + 8 + 15 = 31.

But let me check with an example. If x is 15, then the sides are 8, 8, 15. Let me check if that's a valid triangle.

Yes, because 8 + 8 > 15 (16 > 15), which is true. The other inequalities are trivial.

Alternatively, if x was 16, then 8 + 8 = 16, which is not greater, so it's invalid.

Therefore, 15 is the maximum possible value for x.

But wait, let me think again. Suppose someone says that the third side must be less than the sum of the other two sides, but also greater than the difference. Wait, the difference here is 8 - 8 = 0, so x must be greater than 0. So, x can be 1 to 15.

But maybe there's a mistake here? Let me check with another example. Suppose the two sides are 8 and 8, and the third side is 1. Then, 8 + 1 > 8? 9 > 8, which is true. Similarly, 8 + 8 > 1, which is true. So, that works.

So, the third side can be as small as 1, and as large as 15. Therefore, the maximum perimeter is 8 + 8 + 15 = 31.

But let me check if there's a different interpretation of the problem. The problem says "two sides of a triangle are each 8 units long." So, that means two sides are 8, and the third is something else. So, the triangle is isoceles with two sides of 8.

Therefore, the answer should be 31.

But let me check again. Maybe I made a mistake in the triangle inequality. Let me recall the exact statement.

The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

So, for sides a, b, c:

a + b > c

a + c > b

b + c > a

So, in our case, sides are 8, 8, x.

So:

First inequality: 8 + 8 > x => 16 > x

Second inequality: 8 + x > 8 => x > 0

Third inequality: 8 + x > 8 => same as second, x > 0

Therefore, x must be greater than 0 and less than 16. Since x is a whole number, the maximum value is 15.

Therefore, perimeter is 8 + 8 + 15 = 31.

Therefore, the answer is 31.

But let me check if there's a possibility that the third side could be longer? For example, if the triangle was not isoceles? Wait, no, the problem says two sides are each 8 units long. So, the third side is different.

Alternatively, maybe the problem is in another way? For example, if the third side is also 8, then the triangle is equilateral, but then the perimeter would be 24. But since we are looking for the greatest possible perimeter, 31 is bigger than 24, so 31 is better.

Therefore, I think 31 is the correct answer.

But let me check once more. Suppose x is 15. Then, the sides are 8, 8, 15. Let me check if that's a valid triangle.

Yes, because 8 + 8 > 15 (16 > 15). The other inequalities are 8 + 15 > 8 (23 > 8) and 8 + 15 > 8 (same). So, all inequalities are satisfied.

Alternatively, if x was 16, then 8 + 8 = 16, which is not greater, so it's invalid.

Therefore, 15 is the maximum.

Therefore, the greatest possible perimeter is 8 + 8 + 15 = 31.

I think that's solid. So, the answer is 31.

**Final Answer**
The greatest possible perimeter is \boxed{31}.

To determine the **greatest possible perimeter** of a triangle where **two sides are each 8 units long**, and the **third side has a whole number length**, we must apply the **triangle inequality theorem**.

---

### Step 1: Understand the Triangle Inequality

The triangle inequality theorem states that for any triangle with sides $ a $, $ b $, and $ c $, the following must be true:

$$
a + b > c \\
a + c > b \\
b + c > a
$$

In this case, the triangle has sides of lengths $ 8 $, $ 8 $, and $ x $, where $ x $ is the unknown third side (a whole number). Applying the triangle inequality:

1. $ 8 + 8 > x \Rightarrow 16 > x $
2. $ 8 + x > 8 \Rightarrow x > 0 $
3. $ 8 + x > 8 \Rightarrow x > 0 $

So, combining these, we find that:

$$
0 < x < 16
$$

Since $ x $ must be a **whole number**, the possible values for $ x $ are:

$$
x = 1, 2, 3, \dots, 15
$$

---

### Step 2: Maximize the Perimeter

The perimeter $ P $ of the triangle is:

$$
P = 8 + 8 + x = 16 + x
$$

To **maximize** the perimeter, we need to **maximize** $ x $, subject to the constraint $ x < 16 $. The largest whole number less than 16 is:

$$
x = 15
$$

---

### Step 3: Verify the Triangle Inequality for $ x = 15 $

Check if the triangle with sides $ 8, 8, 15 $ is valid:

- $ 8 + 8 = 16 > 15 $ ✅
- $ 8 + 15 = 23 > 8 $ ✅
- $ 8 + 15 = 23 > 8 $ ✅

All triangle inequalities are satisfied, so this is a valid triangle.

---

### Step 4: Compute the Perimeter

$$
P = 8 + 8 + 15 = 31
$$

---

### Final Answer

$$
\boxed{31}
$$

Solution: The Triangle Inequality says that the sum of the lengths of any two sides must be greater than the length of the third side. That means $8+8=16$ must be greater than the length of the third side. The third side has a whole number length, so the greatest possible length is 15 units. That makes the perimeter $8+8+15=\boxed{31}$ units.


Explanation:
Okay, so I need to find the area of this trapezoid. Let me recall the formula for the area of a trapezoid. I think it's something like the average of the two bases multiplied by the height. Let me check that. Yeah, the formula is $\frac{(base_1 + base_2)}{2} \times height$. Right, so I need to figure out which sides are the bases and what the height is.

Looking at the Asymptote code, it draws a trapezoid with coordinates from (0,0) to (27,0) to (15,9) to (0,9) and back to (0,0). Let me visualize that. So the bottom base is from (0,0) to (27,0), which is 27 units long. Then it goes up to (15,9), then left to (0,9), and back down. So the top base is from (0,9) to (15,9), which is 15 units long. Wait, but the Asymptote code labels some sides. Let me check the labels again.

The Asymptote code labels "5 cm" at (21,4.5) which is NE, so that's probably the non-parallel side? Wait, but the trapezoid has two parallel sides. The bases. The other labels: "5 cm" at (7.5,9), which is the midpoint of the top base, so that must be the length of the top base. Wait, but if the top base is from (0,9) to (15,9), that's 15 units. But the label says 5 cm? Wait, that seems conflicting. Maybe I need to check the coordinates again.

Wait, maybe the coordinates are not to scale? Because the Asymptote code is just for drawing, and the labels are given. Let me think. The Asymptote code draws the trapezoid with coordinates (0,0), (27,0), (15,9), (0,9). So the bottom base is from (0,0) to (27,0), which is 27 units. The top base is from (0,9) to (15,9), which is 15 units. Then the legs are from (27,0) to (15,9) and from (0,9) to (0,0). Wait, but the right angle marks are at (0,9), (0,0), (27,0) and (0,0), (0,9), (15,9). So that means the sides from (0,0) to (0,9) is vertical, and from (0,0) to (27,0) is horizontal. So the trapezoid has right angles at (0,0) and (0,9). Therefore, it's a right trapezoid with two right angles.

But the labels given are: "5 cm" at (21,4.5) NE, which is probably the leg from (27,0) to (15,9). Let me check. The coordinates of that leg are from (27,0) to (15,9). The midpoint of that segment would be at ((27+15)/2, (0+9)/2) = (21, 4.5). So that label is on that leg, which is labeled as 5 cm. Then "5 cm" at (7.5,9), which is the midpoint of the top base from (0,9) to (15,9). So that top base is labeled as 5 cm? But the length from (0,9) to (15,9) is 15 units. Wait, but if the label says 5 cm, maybe the coordinates are not in centimeters? Or maybe the figure is scaled?

Wait, maybe the Asymptote code is just for drawing, and the labels are the actual measurements. So the top base is labeled as 5 cm, the left side is labeled as 3 cm, the bottom base is labeled as 9 cm, and the leg is labeled as 5 cm. Let me check again.

The Asymptote code labels:

- "5 cm" at (21,4.5), NE: this is the leg from (27,0) to (15,9). So that's one of the non-parallel sides.

- "5 cm" at (7.5,9), N: this is the top base, from (0,9) to (15,9). So that's 15 units in the drawing, but labeled as 5 cm. So maybe the drawing is scaled? Or maybe the coordinates are not to scale? Wait, but the other labels:

- "3 cm" at (0,4.5), W: this is the left side from (0,0) to (0,9). The midpoint is (0,4.5), so that's labeled as 3 cm. So that vertical side is 3 cm. But in the coordinates, it's from (0,0) to (0,9), which is 9 units. So if that's 3 cm, then each unit is 0.333... cm? Wait, that seems inconsistent with the other labels.

Wait, the bottom base is labeled "9 cm" at (13.5,0), S. The bottom base is from (0,0) to (27,0), so midpoint is (13.5, 0). So that's labeled as 9 cm. So in the drawing, the bottom base is 27 units, but labeled as 9 cm. Therefore, the scale is 27 units = 9 cm, so 1 unit = 1/3 cm. Therefore, the vertical side from (0,0) to (0,9) is 9 units, which would be 9*(1/3) cm = 3 cm. Which matches the label "3 cm" on the left side. Similarly, the top base is 15 units, which would be 15*(1/3) cm = 5 cm. Which matches the label "5 cm" on the top base. Then the leg from (27,0) to (15,9) is sqrt((27-15)^2 + (0-9)^2) = sqrt(12^2 + 9^2) = sqrt(144 + 81) = sqrt(225) = 15 units. Which would be 15*(1/3) cm = 5 cm. Which matches the label "5 cm" on that leg. So that's consistent. Therefore, the coordinates are scaled such that 1 unit = 1/3 cm. Therefore, the actual lengths are as labeled.

Therefore, to compute the area, we can use the formula for the trapezoid. The two bases are the top and bottom bases. The top base is 5 cm, the bottom base is 9 cm. Wait, but hold on. Wait, the bottom base is labeled as 9 cm, and the top base is labeled as 5 cm. Then the height of the trapezoid is the distance between the two bases. Since it's a right trapezoid, the height is the vertical side, which is 3 cm. Wait, but let me confirm.

In a trapezoid, the height is the perpendicular distance between the two bases. Since it's a right trapezoid with right angles at (0,0) and (0,9), the vertical side from (0,0) to (0,9) is perpendicular to both bases. Therefore, the height is 3 cm. Wait, but the vertical side is labeled as 3 cm. So that's the height. Therefore, the area is ((base1 + base2)/2) * height = ((5 + 9)/2) * 3 = (14/2)*3 = 7*3 = 21 cm². Is that correct?

Wait, but let me check again. Alternatively, maybe I need to use the coordinates to calculate the area? Let me try that. The coordinates are (0,0), (27,0), (15,9), (0,9). If I use the coordinates, the area can be calculated using the shoelace formula. Let me try that.

Shoelace formula: For coordinates (x1,y1), (x2,y2), (x3,y3), (x4,y4), the area is |(x1y2 - x2y1 + x2y3 - x3y2 + x3y4 - x4y3 + x4y1 - x1y4)/2|.

Plugging in the coordinates:

x1=0, y1=0

x2=27, y2=0

x3=15, y3=9

x4=0, y4=9

Calculating:

First term: x1y2 = 0*0 = 0

Second term: -x2y1 = -27*0 = 0

Third term: x2y3 = 27*9 = 243

Fourth term: -x3y2 = -15*0 = 0

Fifth term: x3y4 = 15*9 = 135

Sixth term: -x4y3 = -0*9 = 0

Seventh term: x4y1 = 0*0 = 0

Eighth term: -x1y4 = -0*9 = 0

Adding all terms: 0 + 0 + 243 + 0 + 135 + 0 + 0 + 0 = 378

Take absolute value and divide by 2: |378| / 2 = 189. So the area is 189 square units. But since the scale is 1 unit = 1/3 cm, then the area in cm² would be 189 * (1/3)^2 = 189 * 1/9 = 21 cm². Which matches the previous result. So that's good. Therefore, the area is 21 cm².

Alternatively, since we know the height is 3 cm, and the two bases are 5 cm and 9 cm, then ((5 + 9)/2)*3 = 21 cm². So that's straightforward.

But let me just make sure that the height is indeed 3 cm. The height is the distance between the two bases. Since the trapezoid has right angles at the left side, the vertical side is the height. The vertical side is labeled as 3 cm, so that's the height. Therefore, yes, that's correct.

Alternatively, if I didn't notice the right angles, I could have calculated the height. For example, if the trapezoid is not right-angled, but in this case, it is. So the height is 3 cm.

Wait, but let me think again. The two bases are the top and bottom. The bottom base is 9 cm, the top base is 5 cm. The height is 3 cm. So area is average of the bases times height. So (9 + 5)/2 * 3 = 14/2 * 3 = 7 * 3 = 21. Yep.

Alternatively, if I didn't remember the formula, I could divide the trapezoid into simpler shapes. For example, since it's a right trapezoid, maybe split it into a rectangle and a right triangle?

Looking at the trapezoid: from (0,0) to (27,0) to (15,9) to (0,9). If I imagine the rectangle from (0,0) to (something, 0) to (something, 9) to (0,9). But the top base is only 5 cm, which is from (0,9) to (15,9). Wait, but the bottom base is 9 cm. So maybe the trapezoid can be thought of as a rectangle of 5 cm by 3 cm, plus a right triangle on the side?

Wait, let me think. The bottom base is 9 cm, the top base is 5 cm. The difference between the bases is 9 - 5 = 4 cm. Since it's a right trapezoid, the side that is slanting (from (27,0) to (15,9)) is 5 cm, as labeled. If I imagine extending the top base to meet the bottom base, but maybe that's more complicated.

Alternatively, since the vertical side is 3 cm, and the horizontal side from (0,0) to (27,0) is 9 cm. The top base is 5 cm. So if I consider the trapezoid as a rectangle with length 5 cm and height 3 cm, plus a right triangle on the side. The triangle would have a base of 9 - 5 = 4 cm and height of 3 cm? Wait, but the slant side is 5 cm. Let me check.

If the triangle has base 4 cm and height 3 cm, then the hypotenuse would be 5 cm. Because sqrt(4² + 3²) = 5. Yes! So that matches the label of 5 cm on the slant side. Therefore, the area would be the area of the rectangle plus the area of the triangle. The rectangle is 5 cm * 3 cm = 15 cm². The triangle is (4 * 3)/2 = 6 cm². Total area is 15 + 6 = 21 cm². Same result. Perfect, that confirms it again.

Therefore, regardless of the method, the area is 21 square centimeters. So I think that's the answer.

**Final Answer**
The area of the trapezoid is \boxed{21} square centimeters.

To find the area of the trapezoid, we begin by identifying the key components of the shape and the given measurements.

---

### Step 1: Understand the Shape and Given Measurements

The trapezoid is a **right trapezoid**, meaning it has two right angles. From the Asymptote diagram and the labels:

- The **top base** is labeled **5 cm**.
- The **bottom base** is labeled **9 cm**.
- The **height** (the perpendicular distance between the two bases) is **3 cm**, as it is the vertical side from (0,0) to (0,9), which is labeled as 3 cm.
- The **slanted side** (from (27,0) to (15,9)) is labeled **5 cm**, which matches the hypotenuse of a right triangle with legs 3 cm and 4 cm (since $ \sqrt{3^2 + 4^2} = 5 $).

---

### Step 2: Apply the Trapezoid Area Formula

The formula for the area of a trapezoid is:

$$
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
$$

Substituting the known values:

$$
\text{Area} = \frac{1}{2} \times (5 + 9) \times 3 = \frac{1}{2} \times 14 \times 3 = 7 \times 3 = 21
$$

---

### Step 3: Confirm with an Alternative Method

We can also verify the area by decomposing the trapezoid into a **rectangle** and a **right triangle**:

- The **rectangle** has dimensions 5 cm (base) × 3 cm (height) = 15 cm².
- The **right triangle** has base 4 cm (difference between the two bases) and height 3 cm, so its area is:

$$
\frac{1}{2} \times 4 \times 3 = 6 \text{ cm}^2
$$

Adding both areas:

$$
15 + 6 = 21 \text{ cm}^2
$$

---

### Final Answer

$$
\boxed{21}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of this trapezoid. Let me recall the formula for the area of a trapezoid. I think it's something like the average of the two bases multiplied by the height. Let me check that. Yeah, the formula is $\frac{(base_1 + base_2)}{2} \times height$. Right, so I need to figure out which sides are the bases and what the height is.

Looking at the Asymptote code, it draws a trapezoid with coordinates from (0,0) to (27,0) to (15,9) to (0,9) and back to (0,0). Let me visualize that. So the bottom base is from (0,0) to (27,0), which is 27 units long. Then it goes up to (15,9), then left to (0,9), and back down. So the top base is from (0,9) to (15,9), which is 15 units long. Wait, but the Asymptote code labels some sides. Let me check the labels again.

The Asymptote code labels "5 cm" at (21,4.5) which is NE, so that's probably the non-parallel side? Wait, but the trapezoid has two parallel sides. The bases. The other labels: "5 cm" at (7.5,9), which is the midpoint of the top base, so that must be the length of the top base. Wait, but if the top base is from (0,9) to (15,9), that's 15 units. But the label says 5 cm? Wait, that seems conflicting. Maybe I need to check the coordinates again.

Wait, maybe the coordinates are not to scale? Because the Asymptote code is just for drawing, and the labels are given. Let me think. The Asymptote code draws the trapezoid with coordinates (0,0), (27,0), (15,9), (0,9). So the bottom base is from (0,0) to (27,0), which is 27 units. The top base is from (0,9) to (15,9), which is 15 units. Then the legs are from (27,0) to (15,9) and from (0,9) to (0,0). Wait, but the right angle marks are at (0,9), (0,0), (27,0) and (0,0), (0,9), (15,9). So that means the sides from (0,0) to (0,9) is vertical, and from (0,0) to (27,0) is horizontal. So the trapezoid has right angles at (0,0) and (0,9). Therefore, it's a right trapezoid with two right angles.

But the labels given are: "5 cm" at (21,4.5) NE, which is probably the leg from (27,0) to (15,9). Let me check. The coordinates of that leg are from (27,0) to (15,9). The midpoint of that segment would be at ((27+15)/2, (0+9)/2) = (21, 4.5). So that label is on that leg, which is labeled as 5 cm. Then "5 cm" at (7.5,9), which is the midpoint of the top base from (0,9) to (15,9). So that top base is labeled as 5 cm? But the length from (0,9) to (15,9) is 15 units. Wait, but if the label says 5 cm, maybe the coordinates are not in centimeters? Or maybe the figure is scaled?

Wait, maybe the Asymptote code is just for drawing, and the labels are the actual measurements. So the top base is labeled as 5 cm, the left side is labeled as 3 cm, the bottom base is labeled as 9 cm, and the leg is labeled as 5 cm. Let me check again.

The Asymptote code labels:

- "5 cm" at (21,4.5), NE: this is the leg from (27,0) to (15,9). So that's one of the non-parallel sides.

- "5 cm" at (7.5,9), N: this is the top base, from (0,9) to (15,9). So that's 15 units in the drawing, but labeled as 5 cm. So maybe the drawing is scaled? Or maybe the coordinates are not to scale? Wait, but the other labels:

- "3 cm" at (0,4.5), W: this is the left side from (0,0) to (0,9). The midpoint is (0,4.5), so that's labeled as 3 cm. So that vertical side is 3 cm. But in the coordinates, it's from (0,0) to (0,9), which is 9 units. So if that's 3 cm, then each unit is 0.333... cm? Wait, that seems inconsistent with the other labels.

Wait, the bottom base is labeled "9 cm" at (13.5,0), S. The bottom base is from (0,0) to (27,0), so midpoint is (13.5, 0). So that's labeled as 9 cm. So in the drawing, the bottom base is 27 units, but labeled as 9 cm. Therefore, the scale is 27 units = 9 cm, so 1 unit = 1/3 cm. Therefore, the vertical side from (0,0) to (0,9) is 9 units, which would be 9*(1/3) cm = 3 cm. Which matches the label "3 cm" on the left side. Similarly, the top base is 15 units, which would be 15*(1/3) cm = 5 cm. Which matches the label "5 cm" on the top base. Then the leg from (27,0) to (15,9) is sqrt((27-15)^2 + (0-9)^2) = sqrt(12^2 + 9^2) = sqrt(144 + 81) = sqrt(225) = 15 units. Which would be 15*(1/3) cm = 5 cm. Which matches the label "5 cm" on that leg. So that's consistent. Therefore, the coordinates are scaled such that 1 unit = 1/3 cm. Therefore, the actual lengths are as labeled.

Therefore, to compute the area, we can use the formula for the trapezoid. The two bases are the top and bottom bases. The top base is 5 cm, the bottom base is 9 cm. Wait, but hold on. Wait, the bottom base is labeled as 9 cm, and the top base is labeled as 5 cm. Then the height of the trapezoid is the distance between the two bases. Since it's a right trapezoid, the height is the vertical side, which is 3 cm. Wait, but let me confirm.

In a trapezoid, the height is the perpendicular distance between the two bases. Since it's a right trapezoid with right angles at (0,0) and (0,9), the vertical side from (0,0) to (0,9) is perpendicular to both bases. Therefore, the height is 3 cm. Wait, but the vertical side is labeled as 3 cm. So that's the height. Therefore, the area is ((base1 + base2)/2) * height = ((5 + 9)/2) * 3 = (14/2)*3 = 7*3 = 21 cm². Is that correct?

Wait, but let me check again. Alternatively, maybe I need to use the coordinates to calculate the area? Let me try that. The coordinates are (0,0), (27,0), (15,9), (0,9). If I use the coordinates, the area can be calculated using the shoelace formula. Let me try that.

Shoelace formula: For coordinates (x1,y1), (x2,y2), (x3,y3), (x4,y4), the area is |(x1y2 - x2y1 + x2y3 - x3y2 + x3y4 - x4y3 + x4y1 - x1y4)/2|.

Plugging in the coordinates:

x1=0, y1=0

x2=27, y2=0

x3=15, y3=9

x4=0, y4=9

Calculating:

First term: x1y2 = 0*0 = 0

Second term: -x2y1 = -27*0 = 0

Third term: x2y3 = 27*9 = 243

Fourth term: -x3y2 = -15*0 = 0

Fifth term: x3y4 = 15*9 = 135

Sixth term: -x4y3 = -0*9 = 0

Seventh term: x4y1 = 0*0 = 0

Eighth term: -x1y4 = -0*9 = 0

Adding all terms: 0 + 0 + 243 + 0 + 135 + 0 + 0 + 0 = 378

Take absolute value and divide by 2: |378| / 2 = 189. So the area is 189 square units. But since the scale is 1 unit = 1/3 cm, then the area in cm² would be 189 * (1/3)^2 = 189 * 1/9 = 21 cm². Which matches the previous result. So that's good. Therefore, the area is 21 cm².

Alternatively, since we know the height is 3 cm, and the two bases are 5 cm and 9 cm, then ((5 + 9)/2)*3 = 21 cm². So that's straightforward.

But let me just make sure that the height is indeed 3 cm. The height is the distance between the two bases. Since the trapezoid has right angles at the left side, the vertical side is the height. The vertical side is labeled as 3 cm, so that's the height. Therefore, yes, that's correct.

Alternatively, if I didn't notice the right angles, I could have calculated the height. For example, if the trapezoid is not right-angled, but in this case, it is. So the height is 3 cm.

Wait, but let me think again. The two bases are the top and bottom. The bottom base is 9 cm, the top base is 5 cm. The height is 3 cm. So area is average of the bases times height. So (9 + 5)/2 * 3 = 14/2 * 3 = 7 * 3 = 21. Yep.

Alternatively, if I didn't remember the formula, I could divide the trapezoid into simpler shapes. For example, since it's a right trapezoid, maybe split it into a rectangle and a right triangle?

Looking at the trapezoid: from (0,0) to (27,0) to (15,9) to (0,9). If I imagine the rectangle from (0,0) to (something, 0) to (something, 9) to (0,9). But the top base is only 5 cm, which is from (0,9) to (15,9). Wait, but the bottom base is 9 cm. So maybe the trapezoid can be thought of as a rectangle of 5 cm by 3 cm, plus a right triangle on the side?

Wait, let me think. The bottom base is 9 cm, the top base is 5 cm. The difference between the bases is 9 - 5 = 4 cm. Since it's a right trapezoid, the side that is slanting (from (27,0) to (15,9)) is 5 cm, as labeled. If I imagine extending the top base to meet the bottom base, but maybe that's more complicated.

Alternatively, since the vertical side is 3 cm, and the horizontal side from (0,0) to (27,0) is 9 cm. The top base is 5 cm. So if I consider the trapezoid as a rectangle with length 5 cm and height 3 cm, plus a right triangle on the side. The triangle would have a base of 9 - 5 = 4 cm and height of 3 cm? Wait, but the slant side is 5 cm. Let me check.

If the triangle has base 4 cm and height 3 cm, then the hypotenuse would be 5 cm. Because sqrt(4² + 3²) = 5. Yes! So that matches the label of 5 cm on the slant side. Therefore, the area would be the area of the rectangle plus the area of the triangle. The rectangle is 5 cm * 3 cm = 15 cm². The triangle is (4 * 3)/2 = 6 cm². Total area is 15 + 6 = 21 cm². Same result. Perfect, that confirms it again.

Therefore, regardless of the method, the area is 21 square centimeters. So I think that's the answer.

**Final Answer**
The area of the trapezoid is \boxed{21} square centimeters.

To find the area of the trapezoid, we begin by identifying the key components of the shape and the given measurements.

---

### Step 1: Understand the Shape and Given Measurements

The trapezoid is a **right trapezoid**, meaning it has two right angles. From the Asymptote diagram and the labels:

- The **top base** is labeled **5 cm**.
- The **bottom base** is labeled **9 cm**.
- The **height** (the perpendicular distance between the two bases) is **3 cm**, as it is the vertical side from (0,0) to (0,9), which is labeled as 3 cm.
- The **slanted side** (from (27,0) to (15,9)) is labeled **5 cm**, which matches the hypotenuse of a right triangle with legs 3 cm and 4 cm (since $ \sqrt{3^2 + 4^2} = 5 $).

---

### Step 2: Apply the Trapezoid Area Formula

The formula for the area of a trapezoid is:

$$
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
$$

Substituting the known values:

$$
\text{Area} = \frac{1}{2} \times (5 + 9) \times 3 = \frac{1}{2} \times 14 \times 3 = 7 \times 3 = 21
$$

---

### Step 3: Confirm with an Alternative Method

We can also verify the area by decomposing the trapezoid into a **rectangle** and a **right triangle**:

- The **rectangle** has dimensions 5 cm (base) × 3 cm (height) = 15 cm².
- The **right triangle** has base 4 cm (difference between the two bases) and height 3 cm, so its area is:

$$
\frac{1}{2} \times 4 \times 3 = 6 \text{ cm}^2
$$

Adding both areas:

$$
15 + 6 = 21 \text{ cm}^2
$$

---

### Final Answer

$$
\boxed{21}
$$

Solution:
Draw the dashed line segment in the figure below to divide the trapezoid into a rectangle and a right triangle.  The area of the rectangle is $(5\text{ cm})(3\text{ cm})=15\text{ cm}^2$, and the area of the triangle is $\frac{1}{2}(3\text{ cm})(9\text{ cm}-5\text{ cm})=6\text{ cm}^2$.  Adding the area of the rectangle and the area of the triangle, we find that the area of the trapezoid is $\boxed{21}$ square centimeters. [asy]
defaultpen(linewidth(0.7));
draw((0,0)--(27,0)--(15,9)--(0,9)--cycle);
label("5 cm",(21,4.5),NE);
label("5 cm",(7.5,9),N);
label("3 cm",(0,4.5),W);
label("9 cm",(13.5,0),S);
draw(rightanglemark((0,9),(0,0),(27,0),35));
draw(rightanglemark((0,0),(0,9),(15,9),35));
draw(rightanglemark((15,9),(15,0),(27,0),35));
draw((15,0)--(15,9),linetype("2 4"));
[/asy]