Explanation:
Okay, so I need to find the value of y such that the projection of the vector [2, y, -5] onto the vector [1, -2, 1] is equal to (5/6)*[1, -2, 1]. Hmm, projections... Let me recall how that works.

First, the formula for the projection of a vector **a** onto another vector **b** is given by:

proj_b **a** = ( (**a** · **b**) / ||**b**||² ) * **b**

So, in this case, **a** is [2, y, -5] and **b** is [1, -2, 1]. The projection is given as (5/6)*[1, -2, 1]. Therefore, I can set up the equation using the projection formula and solve for y.

Let me compute the dot product **a** · **b** first. That would be:

(2)(1) + (y)(-2) + (-5)(1) = 2 - 2y - 5 = (-3 - 2y)

Then, the norm squared of **b** is ||**b**||² = (1)² + (-2)² + (1)² = 1 + 4 + 1 = 6.

So plugging into the projection formula:

proj_b **a** = ( (-3 - 2y)/6 ) * [1, -2, 1]

But according to the problem, this projection is equal to (5/6)*[1, -2, 1]. Therefore, we can set the scalar multiples equal to each other because the vectors are scalar multiples of [1, -2, 1].

So:

(-3 - 2y)/6 = 5/6

Since the denominators are the same, we can equate the numerators:

-3 - 2y = 5

Solving for y:

-2y = 5 + 3

-2y = 8

y = 8 / (-2) = -4

Wait, so y is -4? Let me check my steps again to make sure I didn't make a mistake.

First, the dot product: 2*1 is 2, y*(-2) is -2y, -5*1 is -5. So total is 2 - 2y -5 = -3 - 2y. That seems right.

Then ||**b**||² is 1 + 4 + 1 = 6. Correct.

So the scalar multiplier is (-3 - 2y)/6. The projection is that scalar times **b**, which is equal to (5/6)**b**. Therefore, the scalar must be 5/6. Therefore:

(-3 - 2y)/6 = 5/6 => -3 - 2y = 5 => -2y = 8 => y = -4. Seems correct.

But let me verify by plugging y = -4 back into the original vector and computing the projection.

Original vector **a** is [2, -4, -5]. Let's compute the projection.

First, **a** · **b** = 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 - 5 = 5. Wait, that's 5. Then divide by ||**b**||² which is 6. So the scalar is 5/6. Therefore, the projection is (5/6)*[1, -2, 1], which matches the given projection. So that checks out. Therefore, y = -4 is correct.

Wait, but hold on, when I first calculated the dot product with y, I had -3 - 2y. If y = -4, then that would be -3 -2*(-4) = -3 + 8 = 5. Which is correct. So that's exactly what I just did. So yes, that works. Therefore, the answer is y = -4.

But let me think again if there's another way this could be approached, just to confirm.

Alternatively, since the projection is given, maybe we can think about the fact that the projection vector is a scalar multiple of **b**, so the projection vector is (5/6)**b**. Therefore, the projection of **a** onto **b** is (5/6)**b**. Therefore, the component of **a** in the direction of **b** is (5/6)**b**. Therefore, the vector **a** can be written as the projection plus a vector orthogonal to **b**. But maybe that's complicating things. However, since we already solved it through the formula and verified by plugging back in, I think that's sufficient.

Alternatively, maybe using coordinates. Let me think. The projection formula gives us that the projection vector is ( (a · b)/||b||² ) * b. So if we equate that to (5/6)b, then (a · b)/||b||² = 5/6. Therefore, a · b = (5/6)*||b||². Since ||b||² is 6, then a · b = 5/6 * 6 = 5. Therefore, a · b = 5. Which is exactly what we found when we plugged in y = -4. Therefore, solving a · b = 5 gives us the same equation. So that's another way to think about it. So that's consistent.

Therefore, I think my answer is correct. y = -4.

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

To find the value of $ y $ such that the projection of the vector
$$
\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}
$$
onto the vector
$$
\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$
is
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix},
$$
we use the standard formula for the projection of a vector $ \mathbf{a} $ onto another vector $ \mathbf{b} $:

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|^2} \right) \mathbf{b}
$$

---

### Step 1: Compute the dot product $ \mathbf{a} \cdot \mathbf{b} $

Let
$$
\mathbf{a} = \begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Then
$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + y \cdot (-2) + (-5) \cdot 1 = 2 - 2y - 5 = -3 - 2y
$$

---

### Step 2: Compute the squared norm of $ \mathbf{b} $

$$
\|\mathbf{b}\|^2 = 1^2 + (-2)^2 + 1^2 = 1 + 4 + 1 = 6
$$

---

### Step 3: Apply the projection formula

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{-3 - 2y}{6} \right) \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

We are told that this projection equals
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

So we equate the scalar multipliers:

$$
\frac{-3 - 2y}{6} = \frac{5}{6}
$$

Multiply both sides by 6:

$$
-3 - 2y = 5
$$

Solve for $ y $:

$$
-2y = 5 + 3 = 8 \quad \Rightarrow \quad y = \frac{8}{-2} = -4
$$

---

### Step 4: Verify the result

Let’s plug $ y = -4 $ back into the original vector:

$$
\mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ -5 \end{pmatrix}
$$

Compute the dot product:

$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + (-4) \cdot (-2) + (-5) \cdot 1 = 2 + 8 - 5 = 5
$$

Then the projection is:

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{5}{6} \right) \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Which matches the given projection. So the solution is verified.

---

### Final Answer

$$
\boxed{-4}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of y such that the projection of the vector [2, y, -5] onto the vector [1, -2, 1] is equal to (5/6)*[1, -2, 1]. Hmm, projections... Let me recall how that works.

First, the formula for the projection of a vector **a** onto another vector **b** is given by:

proj_b **a** = ( (**a** · **b**) / ||**b**||² ) * **b**

So, in this case, **a** is [2, y, -5] and **b** is [1, -2, 1]. The projection is given as (5/6)*[1, -2, 1]. Therefore, I can set up the equation using the projection formula and solve for y.

Let me compute the dot product **a** · **b** first. That would be:

(2)(1) + (y)(-2) + (-5)(1) = 2 - 2y - 5 = (-3 - 2y)

Then, the norm squared of **b** is ||**b**||² = (1)² + (-2)² + (1)² = 1 + 4 + 1 = 6.

So plugging into the projection formula:

proj_b **a** = ( (-3 - 2y)/6 ) * [1, -2, 1]

But according to the problem, this projection is equal to (5/6)*[1, -2, 1]. Therefore, we can set the scalar multiples equal to each other because the vectors are scalar multiples of [1, -2, 1].

So:

(-3 - 2y)/6 = 5/6

Since the denominators are the same, we can equate the numerators:

-3 - 2y = 5

Solving for y:

-2y = 5 + 3

-2y = 8

y = 8 / (-2) = -4

Wait, so y is -4? Let me check my steps again to make sure I didn't make a mistake.

First, the dot product: 2*1 is 2, y*(-2) is -2y, -5*1 is -5. So total is 2 - 2y -5 = -3 - 2y. That seems right.

Then ||**b**||² is 1 + 4 + 1 = 6. Correct.

So the scalar multiplier is (-3 - 2y)/6. The projection is that scalar times **b**, which is equal to (5/6)**b**. Therefore, the scalar must be 5/6. Therefore:

(-3 - 2y)/6 = 5/6 => -3 - 2y = 5 => -2y = 8 => y = -4. Seems correct.

But let me verify by plugging y = -4 back into the original vector and computing the projection.

Original vector **a** is [2, -4, -5]. Let's compute the projection.

First, **a** · **b** = 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 - 5 = 5. Wait, that's 5. Then divide by ||**b**||² which is 6. So the scalar is 5/6. Therefore, the projection is (5/6)*[1, -2, 1], which matches the given projection. So that checks out. Therefore, y = -4 is correct.

Wait, but hold on, when I first calculated the dot product with y, I had -3 - 2y. If y = -4, then that would be -3 -2*(-4) = -3 + 8 = 5. Which is correct. So that's exactly what I just did. So yes, that works. Therefore, the answer is y = -4.

But let me think again if there's another way this could be approached, just to confirm.

Alternatively, since the projection is given, maybe we can think about the fact that the projection vector is a scalar multiple of **b**, so the projection vector is (5/6)**b**. Therefore, the projection of **a** onto **b** is (5/6)**b**. Therefore, the component of **a** in the direction of **b** is (5/6)**b**. Therefore, the vector **a** can be written as the projection plus a vector orthogonal to **b**. But maybe that's complicating things. However, since we already solved it through the formula and verified by plugging back in, I think that's sufficient.

Alternatively, maybe using coordinates. Let me think. The projection formula gives us that the projection vector is ( (a · b)/||b||² ) * b. So if we equate that to (5/6)b, then (a · b)/||b||² = 5/6. Therefore, a · b = (5/6)*||b||². Since ||b||² is 6, then a · b = 5/6 * 6 = 5. Therefore, a · b = 5. Which is exactly what we found when we plugged in y = -4. Therefore, solving a · b = 5 gives us the same equation. So that's another way to think about it. So that's consistent.

Therefore, I think my answer is correct. y = -4.

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

To find the value of $ y $ such that the projection of the vector
$$
\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}
$$
onto the vector
$$
\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$
is
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix},
$$
we use the standard formula for the projection of a vector $ \mathbf{a} $ onto another vector $ \mathbf{b} $:

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|^2} \right) \mathbf{b}
$$

---

### Step 1: Compute the dot product $ \mathbf{a} \cdot \mathbf{b} $

Let
$$
\mathbf{a} = \begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Then
$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + y \cdot (-2) + (-5) \cdot 1 = 2 - 2y - 5 = -3 - 2y
$$

---

### Step 2: Compute the squared norm of $ \mathbf{b} $

$$
\|\mathbf{b}\|^2 = 1^2 + (-2)^2 + 1^2 = 1 + 4 + 1 = 6
$$

---

### Step 3: Apply the projection formula

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{-3 - 2y}{6} \right) \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

We are told that this projection equals
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

So we equate the scalar multipliers:

$$
\frac{-3 - 2y}{6} = \frac{5}{6}
$$

Multiply both sides by 6:

$$
-3 - 2y = 5
$$

Solve for $ y $:

$$
-2y = 5 + 3 = 8 \quad \Rightarrow \quad y = \frac{8}{-2} = -4
$$

---

### Step 4: Verify the result

Let’s plug $ y = -4 $ back into the original vector:

$$
\mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ -5 \end{pmatrix}
$$

Compute the dot product:

$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + (-4) \cdot (-2) + (-5) \cdot 1 = 2 + 8 - 5 = 5
$$

Then the projection is:

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{5}{6} \right) \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Which matches the given projection. So the solution is verified.

---

### Final Answer

$$
\boxed{-4}
$$

Solution:
The projection of $\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}$ onto $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ is
\[\frac{\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}}{\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \frac{-2y - 3}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.\]Then $-2y - 3 = 5,$ so $y = \boxed{-4}.$


Explanation:
Okay, so I need to simplify this expression: (-k + 4) + (-2 + 3k). Let me think about how to approach this. Alright, first, I remember that when you have parentheses in an expression, you usually need to distribute any signs or coefficients, but in this case, there's just addition between the two parentheses. So maybe I can just remove the parentheses and combine like terms?

Let me write it out step by step. The original expression is (-k + 4) + (-2 + 3k). If I take away the parentheses, it should be -k + 4 - 2 + 3k. Wait, is that right? Let me check. The first part is (-k + 4), so when you add the second part, which is (-2 + 3k), you just add each term. So yes, it's like adding -2 and 3k to the first expression. So that would be -k + 4 - 2 + 3k.

Now, I need to combine like terms. The like terms here are the terms with k and the constant terms. Let me separate them. The terms with k are -k and 3k. The constant terms are 4 and -2.

Starting with the k terms: -k + 3k. Hmm, combining those. Let me think. If I have -1k + 3k, that's like (-1 + 3)k, which is 2k. Okay, so that simplifies to 2k.

Now the constants: 4 - 2. That's straightforward. 4 minus 2 is 2. So putting those together, the simplified expression should be 2k + 2.

Wait, let me double-check my steps. Original expression: (-k + 4) + (-2 + 3k). Remove parentheses: -k + 4 - 2 + 3k. Combine like terms: (-k + 3k) + (4 - 2). Which is 2k + 2. Yeah, that seems right.

But let me try another way to verify. Maybe rearrange the terms. Let's see. If I group the k terms first: (-k + 3k) and then the constants (4 - 2). So again, that's 2k + 2.

Alternatively, maybe I can factor something out? Wait, 2k + 2 can be factored as 2(k + 1), but the question just says to simplify, so maybe 2k + 2 is the simplest form. Unless there's a different way to present it, but I think that's it.

Let me check if I made any sign errors. The original expression has -k and then + (-2). So when you remove the parentheses, it's -k + 4 - 2 + 3k. Yes, because adding a negative 2 is the same as subtracting 2. So that's correct.

Another way to check is to plug in a value for k and see if both the original expression and the simplified one give the same result. Let me pick k = 1.

Original expression: (-1 + 4) + (-2 + 3*1) = (3) + (-2 + 3) = 3 + 1 = 4.

Simplified expression: 2*1 + 2 = 2 + 2 = 4. Okay, same result.

Let me try another value, maybe k = 0.

Original expression: (-0 + 4) + (-2 + 0) = (4) + (-2) = 2.

Simplified expression: 2*0 + 2 = 0 + 2 = 2. Also same.

How about k = 2?

Original: (-2 + 4) + (-2 + 6) = (2) + (4) = 6.

Simplified: 2*2 + 2 = 4 + 2 = 6. Perfect, same result.

So that seems to confirm that the simplification is correct.

Therefore, the simplified form of (-k + 4) + (-2 + 3k) is 2k + 2.

**Final Answer**
The simplified expression is \boxed{2k + 2}.

To simplify the expression $(-k + 4) + (-2 + 3k)$, we follow a systematic approach by first removing the parentheses and then combining like terms.

---

### Step 1: Remove Parentheses

The expression is:

$$
(-k + 4) + (-2 + 3k)
$$

Since there is no multiplication or distribution involved, we can simply remove the parentheses:

$$
-k + 4 - 2 + 3k
$$

---

### Step 2: Combine Like Terms

Now, we identify and combine the like terms:

- **Terms with $k$**: $-k + 3k = 2k$
- **Constant terms**: $4 - 2 = 2$

So, combining these:

$$
2k + 2
$$

---

### Step 3: Final Simplified Expression

The simplified form of the expression is:

$$
\boxed{2k + 2}
$$

[Assistant]
Explanation:
Okay, so I need to simplify this expression: (-k + 4) + (-2 + 3k). Let me think about how to approach this. Alright, first, I remember that when you have parentheses in an expression, you usually need to distribute any signs or coefficients, but in this case, there's just addition between the two parentheses. So maybe I can just remove the parentheses and combine like terms?

Let me write it out step by step. The original expression is (-k + 4) + (-2 + 3k). If I take away the parentheses, it should be -k + 4 - 2 + 3k. Wait, is that right? Let me check. The first part is (-k + 4), so when you add the second part, which is (-2 + 3k), you just add each term. So yes, it's like adding -2 and 3k to the first expression. So that would be -k + 4 - 2 + 3k.

Now, I need to combine like terms. The like terms here are the terms with k and the constant terms. Let me separate them. The terms with k are -k and 3k. The constant terms are 4 and -2.

Starting with the k terms: -k + 3k. Hmm, combining those. Let me think. If I have -1k + 3k, that's like (-1 + 3)k, which is 2k. Okay, so that simplifies to 2k.

Now the constants: 4 - 2. That's straightforward. 4 minus 2 is 2. So putting those together, the simplified expression should be 2k + 2.

Wait, let me double-check my steps. Original expression: (-k + 4) + (-2 + 3k). Remove parentheses: -k + 4 - 2 + 3k. Combine like terms: (-k + 3k) + (4 - 2). Which is 2k + 2. Yeah, that seems right.

But let me try another way to verify. Maybe rearrange the terms. Let's see. If I group the k terms first: (-k + 3k) and then the constants (4 - 2). So again, that's 2k + 2.

Alternatively, maybe I can factor something out? Wait, 2k + 2 can be factored as 2(k + 1), but the question just says to simplify, so maybe 2k + 2 is the simplest form. Unless there's a different way to present it, but I think that's it.

Let me check if I made any sign errors. The original expression has -k and then + (-2). So when you remove the parentheses, it's -k + 4 - 2 + 3k. Yes, because adding a negative 2 is the same as subtracting 2. So that's correct.

Another way to check is to plug in a value for k and see if both the original expression and the simplified one give the same result. Let me pick k = 1.

Original expression: (-1 + 4) + (-2 + 3*1) = (3) + (-2 + 3) = 3 + 1 = 4.

Simplified expression: 2*1 + 2 = 2 + 2 = 4. Okay, same result.

Let me try another value, maybe k = 0.

Original expression: (-0 + 4) + (-2 + 0) = (4) + (-2) = 2.

Simplified expression: 2*0 + 2 = 0 + 2 = 2. Also same.

How about k = 2?

Original: (-2 + 4) + (-2 + 6) = (2) + (4) = 6.

Simplified: 2*2 + 2 = 4 + 2 = 6. Perfect, same result.

So that seems to confirm that the simplification is correct.

Therefore, the simplified form of (-k + 4) + (-2 + 3k) is 2k + 2.

**Final Answer**
The simplified expression is \boxed{2k + 2}.

To simplify the expression $(-k + 4) + (-2 + 3k)$, we follow a systematic approach by first removing the parentheses and then combining like terms.

---

### Step 1: Remove Parentheses

The expression is:

$$
(-k + 4) + (-2 + 3k)
$$

Since there is no multiplication or distribution involved, we can simply remove the parentheses:

$$
-k + 4 - 2 + 3k
$$

---

### Step 2: Combine Like Terms

Now, we identify and combine the like terms:

- **Terms with $k$**: $-k + 3k = 2k$
- **Constant terms**: $4 - 2 = 2$

So, combining these:

$$
2k + 2
$$

---

### Step 3: Final Simplified Expression

The simplified form of the expression is:

$$
\boxed{2k + 2}
$$

Solution: We have $(-k+4) + (-2+3k) = -k + 4 -2 + 3k = \boxed{2k+2}$.


Explanation:
Okay, so I need to find the horizontal asymptote of the function y = (4x³ + 2x - 4) / (3x³ - 2x² + 5x - 1). Hmm, horizontal asymptotes... I remember they have something to do with the behavior of the function as x approaches positive or negative infinity. Let me think.

First, I recall that for rational functions, which are ratios of polynomials, the horizontal asymptote depends on the degrees of the numerator and the denominator. The degree is the highest exponent of x in each polynomial. Let me check the degrees here.

The numerator is 4x³ + 2x - 4. The highest power is x³, so the degree is 3. The denominator is 3x³ - 2x² + 5x - 1. Again, the highest power is x³, so the degree is also 3.

So when the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 3. Therefore, the horizontal asymptote should be y = 4/3.

Wait, but let me make sure I'm not missing anything. Sometimes there are cases where the degrees are different, but here they are the same. Let me verify this with another method just to be safe.

Another way to find horizontal asymptotes is to divide the numerator and denominator by the highest power of x in the denominator. In this case, the highest power is x³. So let me do that.

Dividing numerator and denominator by x³:

Numerator becomes (4x³ + 2x - 4)/x³ = 4 + 2/x² - 4/x³

Denominator becomes (3x³ - 2x² + 5x - 1)/x³ = 3 - 2/x + 5/x² - 1/x³

So the function simplifies to [4 + 2/x² - 4/x³] / [3 - 2/x + 5/x² - 1/x³]

Now, as x approaches infinity or negative infinity, the terms with 1/x², 1/x³, etc., will approach zero. Therefore, the expression approaches 4 / 3.

Therefore, the horizontal asymptote is y = 4/3.

But wait, let me check if there are any other horizontal asymptotes. For example, sometimes if the degrees are different, you might have different asymptotes as x approaches positive and negative infinity, but in this case, since the degrees are the same, it should be the same horizontal asymptote for both directions.

Let me also think about the possibility of oblique asymptotes, but no, oblique asymptotes occur when the degree of the numerator is exactly one more than the denominator. Here, both are degree 3, so there's no oblique asymptote.

Another thing to check: maybe there's a mistake in calculating the leading coefficients? Let me confirm. The leading term of the numerator is 4x³, and the leading term of the denominator is 3x³. So the ratio is indeed 4/3.

Alternatively, if I plug in a very large value of x, say x = 1000, then the numerator would be approximately 4*(1000)^3, and the denominator would be approximately 3*(1000)^3, so the ratio is roughly 4/3. Similarly, for x approaching negative infinity, the same thing happens because the x³ term dominates, and even though x is negative, the cube of a negative number is negative, but since both numerator and denominator have x³ terms, their ratio would still be positive 4/3.

Therefore, I think the horizontal asymptote is definitely y = 4/3.

But just to be thorough, let me check if there's any possibility that the function could have a horizontal asymptote at a different value. For example, if there was a cancellation in the numerator and denominator, but in this case, the numerator and denominator are both cubic polynomials. Let me check if they have any common factors.

To check for common factors, I would need to factor both numerator and denominator. Let me try factoring the numerator: 4x³ + 2x - 4. Hmm, factoring cubics can be tricky. Let me see if there are any rational roots using the Rational Root Theorem. The possible rational roots would be factors of the constant term over factors of the leading coefficient. So possible roots are ±1, ±2, ±4, ±1/2, ±1/4. Let me test x=1: 4(1)^3 + 2(1) -4 = 4 + 2 -4 = 2 ≠ 0. x=-1: 4(-1)^3 + 2(-1) -4 = -4 -2 -4 = -10 ≠ 0. x=2: 4(8) + 4 -4 = 32 + 0 = 32 ≠ 0. x=1/2: 4*(1/8) + 2*(1/2) -4 = 0.5 + 1 -4 = -2.5 ≠ 0. x=-1/2: 4*(-1/2)^3 + 2*(-1/2) -4 = 4*(-1/8) -1 -4 = -0.5 -1 -4 = -5.5 ≠ 0. Similarly, x=4 would be way too big. So it seems like the numerator doesn't have any rational roots. Maybe it's irreducible?

Similarly, let me check the denominator: 3x³ - 2x² + 5x -1. Again, possible rational roots are ±1, ±1/3. Let me test x=1: 3 - 2 + 5 -1 = 5 ≠ 0. x=-1: -3 - 2 -5 -1 = -11 ≠ 0. x=1/3: 3*(1/27) - 2*(1/9) + 5*(1/3) -1 = (1/9) - (2/9) + (5/3) -1 = (-1/9) + (5/3 - 1) = (-1/9) + (2/3) = (-1/9 + 6/9) = 5/9 ≠ 0. x=-1/3: 3*(-1/3)^3 - 2*(-1/3)^2 + 5*(-1/3) -1 = 3*(-1/27) - 2*(1/9) -5/3 -1 = (-1/9) - 2/9 -5/3 -1 = (-3/9) -5/3 -1 = (-1/3) -5/3 -1 = (-6/3) -1 = -2 -1 = -3 ≠ 0. So denominator also doesn't have any rational roots.

Therefore, since neither numerator nor denominator can be factored easily, there are no common factors between them. Therefore, there's no cancellation, and the horizontal asymptote is indeed y = 4/3.

Wait, but just to make sure, maybe I should check the limit as x approaches infinity. Let me compute the limit of y as x approaches infinity.

lim(x→∞) [4x³ + 2x -4]/[3x³ -2x² +5x -1].

Divide numerator and denominator by x³:

lim(x→∞) [4 + 2/x² -4/x³]/[3 - 2/x +5/x² -1/x³]

As x approaches infinity, the terms with 1/x², 1/x³, etc., go to zero. Therefore, the limit becomes (4 + 0 - 0)/(3 - 0 + 0 - 0) = 4/3.

Similarly, as x approaches negative infinity, the same thing happens. The terms with 1/x², etc., still go to zero, and the leading terms are 4x³ / 3x³ = 4/3.

Therefore, the horizontal asymptote is y = 4/3.

I think that's solid. I don't see any mistakes in my reasoning. So the answer should be y = 4/3.

**Final Answer**
The horizontal asymptote occurs at \boxed{\dfrac{4}{3}}.

To determine the horizontal asymptote of the function
$$
y = \frac{4x^3 + 2x - 4}{3x^3 - 2x^2 + 5x - 1},
$$
we analyze the behavior of the function as $ x \to \infty $ and $ x \to -\infty $.

---

### Step 1: Identify the Degrees of the Polynomials

- The **numerator** is a cubic polynomial: $ 4x^3 + 2x - 4 $, so its degree is **3**.
- The **denominator** is also a cubic polynomial: $ 3x^3 - 2x^2 + 5x - 1 $, so its degree is **3**.

Since the degrees of the numerator and denominator are equal, the function has a **horizontal asymptote**.

---

### Step 2: Use the Rule for Horizontal Asymptotes

For a rational function of the form
$$
y = \frac{a_n x^n + \cdots}{b_m x^m + \cdots},
$$
where the degrees of the numerator and denominator are both $ n $, the horizontal asymptote is given by
$$
y = \frac{a_n}{b_m},
$$
where $ a_n $ and $ b_m $ are the leading coefficients of the numerator and denominator, respectively.

- Leading coefficient of the numerator: $ 4 $
- Leading coefficient of the denominator: $ 3 $

Thus, the horizontal asymptote is
$$
y = \frac{4}{3}.
$$

---

### Step 3: Confirm with Limit Analysis

To verify, we can divide both the numerator and denominator by the highest power of $ x $, which is $ x^3 $:

$$
y = \frac{4 + \frac{2}{x^2} - \frac{4}{x^3}}{3 - \frac{2}{x} + \frac{5}{x^2} - \frac{1}{x^3}}.
$$

As $ x \to \infty $ or $ x \to -\infty $, the terms involving $ \frac{1}{x} $, $ \frac{1}{x^2} $, and $ \frac{1}{x^3} $ all approach 0. Therefore, the function approaches:

$$
\frac{4}{3}.
$$

This confirms that the horizontal asymptote is indeed $ y = \frac{4}{3} $.

---

### Final Answer

$$
\boxed{\dfrac{4}{3}}
$$

[Assistant]
Explanation:
Okay, so I need to find the horizontal asymptote of the function y = (4x³ + 2x - 4) / (3x³ - 2x² + 5x - 1). Hmm, horizontal asymptotes... I remember they have something to do with the behavior of the function as x approaches positive or negative infinity. Let me think.

First, I recall that for rational functions, which are ratios of polynomials, the horizontal asymptote depends on the degrees of the numerator and the denominator. The degree is the highest exponent of x in each polynomial. Let me check the degrees here.

The numerator is 4x³ + 2x - 4. The highest power is x³, so the degree is 3. The denominator is 3x³ - 2x² + 5x - 1. Again, the highest power is x³, so the degree is also 3.

So when the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 3. Therefore, the horizontal asymptote should be y = 4/3.

Wait, but let me make sure I'm not missing anything. Sometimes there are cases where the degrees are different, but here they are the same. Let me verify this with another method just to be safe.

Another way to find horizontal asymptotes is to divide the numerator and denominator by the highest power of x in the denominator. In this case, the highest power is x³. So let me do that.

Dividing numerator and denominator by x³:

Numerator becomes (4x³ + 2x - 4)/x³ = 4 + 2/x² - 4/x³

Denominator becomes (3x³ - 2x² + 5x - 1)/x³ = 3 - 2/x + 5/x² - 1/x³

So the function simplifies to [4 + 2/x² - 4/x³] / [3 - 2/x + 5/x² - 1/x³]

Now, as x approaches infinity or negative infinity, the terms with 1/x², 1/x³, etc., will approach zero. Therefore, the expression approaches 4 / 3.

Therefore, the horizontal asymptote is y = 4/3.

But wait, let me check if there are any other horizontal asymptotes. For example, sometimes if the degrees are different, you might have different asymptotes as x approaches positive and negative infinity, but in this case, since the degrees are the same, it should be the same horizontal asymptote for both directions.

Let me also think about the possibility of oblique asymptotes, but no, oblique asymptotes occur when the degree of the numerator is exactly one more than the denominator. Here, both are degree 3, so there's no oblique asymptote.

Another thing to check: maybe there's a mistake in calculating the leading coefficients? Let me confirm. The leading term of the numerator is 4x³, and the leading term of the denominator is 3x³. So the ratio is indeed 4/3.

Alternatively, if I plug in a very large value of x, say x = 1000, then the numerator would be approximately 4*(1000)^3, and the denominator would be approximately 3*(1000)^3, so the ratio is roughly 4/3. Similarly, for x approaching negative infinity, the same thing happens because the x³ term dominates, and even though x is negative, the cube of a negative number is negative, but since both numerator and denominator have x³ terms, their ratio would still be positive 4/3.

Therefore, I think the horizontal asymptote is definitely y = 4/3.

But just to be thorough, let me check if there's any possibility that the function could have a horizontal asymptote at a different value. For example, if there was a cancellation in the numerator and denominator, but in this case, the numerator and denominator are both cubic polynomials. Let me check if they have any common factors.

To check for common factors, I would need to factor both numerator and denominator. Let me try factoring the numerator: 4x³ + 2x - 4. Hmm, factoring cubics can be tricky. Let me see if there are any rational roots using the Rational Root Theorem. The possible rational roots would be factors of the constant term over factors of the leading coefficient. So possible roots are ±1, ±2, ±4, ±1/2, ±1/4. Let me test x=1: 4(1)^3 + 2(1) -4 = 4 + 2 -4 = 2 ≠ 0. x=-1: 4(-1)^3 + 2(-1) -4 = -4 -2 -4 = -10 ≠ 0. x=2: 4(8) + 4 -4 = 32 + 0 = 32 ≠ 0. x=1/2: 4*(1/8) + 2*(1/2) -4 = 0.5 + 1 -4 = -2.5 ≠ 0. x=-1/2: 4*(-1/2)^3 + 2*(-1/2) -4 = 4*(-1/8) -1 -4 = -0.5 -1 -4 = -5.5 ≠ 0. Similarly, x=4 would be way too big. So it seems like the numerator doesn't have any rational roots. Maybe it's irreducible?

Similarly, let me check the denominator: 3x³ - 2x² + 5x -1. Again, possible rational roots are ±1, ±1/3. Let me test x=1: 3 - 2 + 5 -1 = 5 ≠ 0. x=-1: -3 - 2 -5 -1 = -11 ≠ 0. x=1/3: 3*(1/27) - 2*(1/9) + 5*(1/3) -1 = (1/9) - (2/9) + (5/3) -1 = (-1/9) + (5/3 - 1) = (-1/9) + (2/3) = (-1/9 + 6/9) = 5/9 ≠ 0. x=-1/3: 3*(-1/3)^3 - 2*(-1/3)^2 + 5*(-1/3) -1 = 3*(-1/27) - 2*(1/9) -5/3 -1 = (-1/9) - 2/9 -5/3 -1 = (-3/9) -5/3 -1 = (-1/3) -5/3 -1 = (-6/3) -1 = -2 -1 = -3 ≠ 0. So denominator also doesn't have any rational roots.

Therefore, since neither numerator nor denominator can be factored easily, there are no common factors between them. Therefore, there's no cancellation, and the horizontal asymptote is indeed y = 4/3.

Wait, but just to make sure, maybe I should check the limit as x approaches infinity. Let me compute the limit of y as x approaches infinity.

lim(x→∞) [4x³ + 2x -4]/[3x³ -2x² +5x -1].

Divide numerator and denominator by x³:

lim(x→∞) [4 + 2/x² -4/x³]/[3 - 2/x +5/x² -1/x³]

As x approaches infinity, the terms with 1/x², 1/x³, etc., go to zero. Therefore, the limit becomes (4 + 0 - 0)/(3 - 0 + 0 - 0) = 4/3.

Similarly, as x approaches negative infinity, the same thing happens. The terms with 1/x², etc., still go to zero, and the leading terms are 4x³ / 3x³ = 4/3.

Therefore, the horizontal asymptote is y = 4/3.

I think that's solid. I don't see any mistakes in my reasoning. So the answer should be y = 4/3.

**Final Answer**
The horizontal asymptote occurs at \boxed{\dfrac{4}{3}}.

To determine the horizontal asymptote of the function
$$
y = \frac{4x^3 + 2x - 4}{3x^3 - 2x^2 + 5x - 1},
$$
we analyze the behavior of the function as $ x \to \infty $ and $ x \to -\infty $.

---

### Step 1: Identify the Degrees of the Polynomials

- The **numerator** is a cubic polynomial: $ 4x^3 + 2x - 4 $, so its degree is **3**.
- The **denominator** is also a cubic polynomial: $ 3x^3 - 2x^2 + 5x - 1 $, so its degree is **3**.

Since the degrees of the numerator and denominator are equal, the function has a **horizontal asymptote**.

---

### Step 2: Use the Rule for Horizontal Asymptotes

For a rational function of the form
$$
y = \frac{a_n x^n + \cdots}{b_m x^m + \cdots},
$$
where the degrees of the numerator and denominator are both $ n $, the horizontal asymptote is given by
$$
y = \frac{a_n}{b_m},
$$
where $ a_n $ and $ b_m $ are the leading coefficients of the numerator and denominator, respectively.

- Leading coefficient of the numerator: $ 4 $
- Leading coefficient of the denominator: $ 3 $

Thus, the horizontal asymptote is
$$
y = \frac{4}{3}.
$$

---

### Step 3: Confirm with Limit Analysis

To verify, we can divide both the numerator and denominator by the highest power of $ x $, which is $ x^3 $:

$$
y = \frac{4 + \frac{2}{x^2} - \frac{4}{x^3}}{3 - \frac{2}{x} + \frac{5}{x^2} - \frac{1}{x^3}}.
$$

As $ x \to \infty $ or $ x \to -\infty $, the terms involving $ \frac{1}{x} $, $ \frac{1}{x^2} $, and $ \frac{1}{x^3} $ all approach 0. Therefore, the function approaches:

$$
\frac{4}{3}.
$$

This confirms that the horizontal asymptote is indeed $ y = \frac{4}{3} $.

---

### Final Answer

$$
\boxed{\dfrac{4}{3}}
$$

Solution:
When the degrees of the numerator and denominator are the same in a rational function, the horizontal asymptote is the coefficient of the highest degree in the numerator divided by the coefficient of the highest degree in the denominator. To see this, divide the numerator and denominator by $x^3$ to write the expression as  \[
\frac{4+\frac{2}{x^2}-\frac{4}{x^3}}{3-\frac{2}{x}+\frac{5}{x^2}-\frac{1}{x^3}}
\]As $x\to\infty$ or $x\to-\infty$, the terms involving $x$ approach 0, which means that the whole expression approaches 4/3. Therefore, there is only one horizontal asymptote, and it is at $y=\boxed{\frac43}$.


Explanation:
Okay, so I need to figure out how many elements are in the intersection of two sets: the set of all prime numbers less than 30 and the set of all odd numbers greater than zero. Let me start by recalling what each of these sets contains and then find their common elements.

First, let me define the two sets clearly.

The first set is all prime numbers less than 30. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So I need to list all primes less than 30. Let me remember the primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Wait, let me check if I missed any. Starting from 2:

- 2 is prime
- 3 is prime
- 4 is not prime (divisible by 2)
- 5 is prime
- 6 is not
- 7 is prime
- 8 not
- 9 not (divisible by 3)
- 10 not
- 11 prime
- 12 not
- 13 prime
- 14 not
- 15 not
- 16 not
- 17 prime
- 18 not
- 19 prime
- 20 not
- 21 not
- 22 not
- 23 prime
- 24 not
- 25 not (5 squared)
- 26 not
- 27 not
- 28 not
- 29 prime
- 30 is not, since we're looking for less than 30.

So the primes less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them. That's 10 primes. Wait, let me count again:

1. 2
2. 3
3. 5
4. 7
5. 11
6. 13
7. 17
8. 19
9. 23
10. 29

Yes, 10 primes. Okay, so the first set has 10 elements.

Now the second set is all odd numbers greater than zero. So odd numbers are numbers that are not divisible by 2. The set of all odd numbers greater than zero would be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, etc. But since we are looking for the intersection with the primes less than 30, we don't need to consider numbers beyond 29. However, the second set is infinite, but since we are intersecting with the first set (which is finite), the intersection will be the primes less than 30 that are also odd numbers greater than zero.

But wait, the second set is all odd numbers greater than zero. So, the intersection would be all numbers that are both prime less than 30 and odd numbers greater than zero. However, note that the only even prime number is 2. All other primes are odd. Therefore, if we take the set of primes less than 30, and remove the even prime (which is 2), then the rest are odd. Therefore, the intersection of the two sets would be all primes less than 30 except 2. But let me verify this.

Alternatively, since the second set includes all odd numbers greater than zero, which includes all odd primes, but also includes composite odd numbers. However, the first set is only primes. Therefore, the intersection would be the primes that are odd numbers greater than zero. Since primes except 2 are odd, so the intersection is all primes less than 30 except 2. Therefore, the number of elements in the intersection would be the number of primes less than 30 minus 1 (for the prime 2). Since there are 10 primes less than 30, subtracting 1 gives 9. Therefore, the answer is 9?

Wait, but let me check again. Let me list the primes less than 30 and see which ones are odd. The primes less than 30 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Now, which of these are odd? Well, 2 is even, so the rest are odd. So 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 9 numbers. Therefore, the intersection is these 9 numbers. Therefore, the answer is 9.

But let me make sure that the second set is indeed all odd numbers greater than zero. So, the second set includes numbers like 1, 3, 5, 7,... So, 1 is in the second set. Is 1 considered here? The problem says "the set of all the odd numbers greater than zero." So, greater than zero, so starting from 1. But 1 is not a prime number. Therefore, the intersection would be the primes less than 30 that are also in the set of odd numbers greater than zero. Since primes are greater than 1, so all primes are greater than 1, so they are in the set of odd numbers greater than zero, except for 2. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

But let me check if 1 is included in the second set. The problem says "the set of all the odd numbers greater than zero." So, numbers greater than zero, so starting at 1. So 1 is included. However, 1 is not prime. Therefore, the intersection would be primes less than 30 that are odd. Since primes except 2 are odd, so the intersection is primes less than 30 except 2. Therefore, 9 elements.

Alternatively, maybe I can think of the intersection as the set of numbers that are both prime less than 30 and odd numbers greater than zero. So, numbers that are primes less than 30 and also odd and greater than zero. Since all primes except 2 are odd, and all primes are greater than zero. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

But let me check if there's any other number that is in both sets. For example, is there a prime number less than 30 that is not odd? Only 2. So, the only prime that is even is 2. Therefore, the rest are odd. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

Alternatively, maybe I can list the elements of both sets and see their intersection.

First set (primes less than 30): {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Second set (odd numbers greater than zero): {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, ...}

So the intersection would be numbers that are in both sets. So, looking at the first set, which numbers are also in the second set? Well, all numbers in the first set except 2 are odd, so they are in the second set. However, 2 is not in the second set because it's even. Therefore, the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}, which has 9 elements.

Therefore, the answer is 9.

But let me check if there's any other number in the first set that is not odd. Well, the only even prime is 2, so no others. Therefore, the intersection is indeed 9 elements.

Alternatively, maybe I can think of the second set as all odd numbers greater than zero. So, the intersection is the set of primes less than 30 that are odd. Since primes are either 2 or odd, so the intersection is primes less than 30 minus {2}. Therefore, 10 - 1 = 9.

Therefore, I think the answer is 9.

But just to be thorough, let me list the numbers in the intersection:

From the first set: 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them:

1. 3
2. 5
3. 7
4. 11
5. 13
6. 17
7. 19
8. 23
9. 29

Yes, that's 9 numbers. Therefore, the intersection has 9 elements.

Therefore, the answer is 9.

But let me check once more if I made a mistake in counting the primes less than 30. Let me list them again:

Primes less than 30:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Yes, that's 10 primes. So removing 2, we have 9 primes left. Therefore, the intersection is 9 elements.

Alternatively, if I think of the set of odd numbers greater than zero, which includes 1, 3, 5,... So, the primes in that set would be all primes except 2. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

Therefore, I think the answer is 9.

But just to make sure, let me check if there's any other number that might be in both sets. For example, is 1 in the first set? No, because 1 is not a prime. So, the intersection is only the primes that are odd. So, yes, 9 elements.

Therefore, I think the answer is 9.

**Final Answer**
The number of elements in the intersection is \boxed{9}.

To determine how many elements are in the intersection of the set of all prime numbers less than 30 and the set of all odd numbers greater than zero, we proceed step by step.

---

### Step 1: Identify the First Set – Prime Numbers Less Than 30

A **prime number** is a natural number greater than 1 that has no positive divisors other than 1 and itself. Listing all such numbers less than 30:

$$
\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **10 elements**.

---

### Step 2: Identify the Second Set – Odd Numbers Greater Than Zero

An **odd number** is any integer that is not divisible by 2. The set of all odd numbers greater than zero includes:

$$
\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, \dots\}
$$

This is an **infinite** set, but since we are only interested in the intersection with the first set (which is finite), we only need to consider the odd numbers that are also in the first set.

---

### Step 3: Find the Intersection

We are looking for numbers that are **both**:

- Prime numbers less than 30, and
- Odd numbers greater than zero.

From the first set, the only even prime is **2**. All other primes are odd. Therefore, the intersection will consist of all the primes less than 30 **except** 2.

So, the intersection is:

$$
\{3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **9 elements**.

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many elements are in the intersection of two sets: the set of all prime numbers less than 30 and the set of all odd numbers greater than zero. Let me start by recalling what each of these sets contains and then find their common elements.

First, let me define the two sets clearly.

The first set is all prime numbers less than 30. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So I need to list all primes less than 30. Let me remember the primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Wait, let me check if I missed any. Starting from 2:

- 2 is prime
- 3 is prime
- 4 is not prime (divisible by 2)
- 5 is prime
- 6 is not
- 7 is prime
- 8 not
- 9 not (divisible by 3)
- 10 not
- 11 prime
- 12 not
- 13 prime
- 14 not
- 15 not
- 16 not
- 17 prime
- 18 not
- 19 prime
- 20 not
- 21 not
- 22 not
- 23 prime
- 24 not
- 25 not (5 squared)
- 26 not
- 27 not
- 28 not
- 29 prime
- 30 is not, since we're looking for less than 30.

So the primes less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them. That's 10 primes. Wait, let me count again:

1. 2
2. 3
3. 5
4. 7
5. 11
6. 13
7. 17
8. 19
9. 23
10. 29

Yes, 10 primes. Okay, so the first set has 10 elements.

Now the second set is all odd numbers greater than zero. So odd numbers are numbers that are not divisible by 2. The set of all odd numbers greater than zero would be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, etc. But since we are looking for the intersection with the primes less than 30, we don't need to consider numbers beyond 29. However, the second set is infinite, but since we are intersecting with the first set (which is finite), the intersection will be the primes less than 30 that are also odd numbers greater than zero.

But wait, the second set is all odd numbers greater than zero. So, the intersection would be all numbers that are both prime less than 30 and odd numbers greater than zero. However, note that the only even prime number is 2. All other primes are odd. Therefore, if we take the set of primes less than 30, and remove the even prime (which is 2), then the rest are odd. Therefore, the intersection of the two sets would be all primes less than 30 except 2. But let me verify this.

Alternatively, since the second set includes all odd numbers greater than zero, which includes all odd primes, but also includes composite odd numbers. However, the first set is only primes. Therefore, the intersection would be the primes that are odd numbers greater than zero. Since primes except 2 are odd, so the intersection is all primes less than 30 except 2. Therefore, the number of elements in the intersection would be the number of primes less than 30 minus 1 (for the prime 2). Since there are 10 primes less than 30, subtracting 1 gives 9. Therefore, the answer is 9?

Wait, but let me check again. Let me list the primes less than 30 and see which ones are odd. The primes less than 30 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Now, which of these are odd? Well, 2 is even, so the rest are odd. So 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 9 numbers. Therefore, the intersection is these 9 numbers. Therefore, the answer is 9.

But let me make sure that the second set is indeed all odd numbers greater than zero. So, the second set includes numbers like 1, 3, 5, 7,... So, 1 is in the second set. Is 1 considered here? The problem says "the set of all the odd numbers greater than zero." So, greater than zero, so starting from 1. But 1 is not a prime number. Therefore, the intersection would be the primes less than 30 that are also in the set of odd numbers greater than zero. Since primes are greater than 1, so all primes are greater than 1, so they are in the set of odd numbers greater than zero, except for 2. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

But let me check if 1 is included in the second set. The problem says "the set of all the odd numbers greater than zero." So, numbers greater than zero, so starting at 1. So 1 is included. However, 1 is not prime. Therefore, the intersection would be primes less than 30 that are odd. Since primes except 2 are odd, so the intersection is primes less than 30 except 2. Therefore, 9 elements.

Alternatively, maybe I can think of the intersection as the set of numbers that are both prime less than 30 and odd numbers greater than zero. So, numbers that are primes less than 30 and also odd and greater than zero. Since all primes except 2 are odd, and all primes are greater than zero. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

But let me check if there's any other number that is in both sets. For example, is there a prime number less than 30 that is not odd? Only 2. So, the only prime that is even is 2. Therefore, the rest are odd. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

Alternatively, maybe I can list the elements of both sets and see their intersection.

First set (primes less than 30): {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Second set (odd numbers greater than zero): {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, ...}

So the intersection would be numbers that are in both sets. So, looking at the first set, which numbers are also in the second set? Well, all numbers in the first set except 2 are odd, so they are in the second set. However, 2 is not in the second set because it's even. Therefore, the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}, which has 9 elements.

Therefore, the answer is 9.

But let me check if there's any other number in the first set that is not odd. Well, the only even prime is 2, so no others. Therefore, the intersection is indeed 9 elements.

Alternatively, maybe I can think of the second set as all odd numbers greater than zero. So, the intersection is the set of primes less than 30 that are odd. Since primes are either 2 or odd, so the intersection is primes less than 30 minus {2}. Therefore, 10 - 1 = 9.

Therefore, I think the answer is 9.

But just to be thorough, let me list the numbers in the intersection:

From the first set: 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them:

1. 3
2. 5
3. 7
4. 11
5. 13
6. 17
7. 19
8. 23
9. 29

Yes, that's 9 numbers. Therefore, the intersection has 9 elements.

Therefore, the answer is 9.

But let me check once more if I made a mistake in counting the primes less than 30. Let me list them again:

Primes less than 30:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Yes, that's 10 primes. So removing 2, we have 9 primes left. Therefore, the intersection is 9 elements.

Alternatively, if I think of the set of odd numbers greater than zero, which includes 1, 3, 5,... So, the primes in that set would be all primes except 2. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

Therefore, I think the answer is 9.

But just to make sure, let me check if there's any other number that might be in both sets. For example, is 1 in the first set? No, because 1 is not a prime. So, the intersection is only the primes that are odd. So, yes, 9 elements.

Therefore, I think the answer is 9.

**Final Answer**
The number of elements in the intersection is \boxed{9}.

To determine how many elements are in the intersection of the set of all prime numbers less than 30 and the set of all odd numbers greater than zero, we proceed step by step.

---

### Step 1: Identify the First Set – Prime Numbers Less Than 30

A **prime number** is a natural number greater than 1 that has no positive divisors other than 1 and itself. Listing all such numbers less than 30:

$$
\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **10 elements**.

---

### Step 2: Identify the Second Set – Odd Numbers Greater Than Zero

An **odd number** is any integer that is not divisible by 2. The set of all odd numbers greater than zero includes:

$$
\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, \dots\}
$$

This is an **infinite** set, but since we are only interested in the intersection with the first set (which is finite), we only need to consider the odd numbers that are also in the first set.

---

### Step 3: Find the Intersection

We are looking for numbers that are **both**:

- Prime numbers less than 30, and
- Odd numbers greater than zero.

From the first set, the only even prime is **2**. All other primes are odd. Therefore, the intersection will consist of all the primes less than 30 **except** 2.

So, the intersection is:

$$
\{3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **9 elements**.

---

### Final Answer

$$
\boxed{9}
$$

Solution: In other words, we're looking for the number of positive odd prime numbers less than 30. We go through all odd numbers less than 30 and note how many of them are prime. We get that 3, 5, 7, 11, 13, 17, 19, 23, and 29 are all of the positive odd prime numbers less than 30, a total of $\boxed{9}$ elements in the intersection.


Explanation:
Okay, so I need to find the value of this expression: (26² - 24² - 10)² - 10². Hmm, let me break this down step by step. First, maybe I should compute the parts inside the parentheses first. Let me recall some algebraic identities that might help simplify things.

Wait, there's a difference of squares formula: a² - b² = (a - b)(a + b). Maybe I can use that here. Let me check if that applies to 26² - 24². Yes, that's exactly a difference of squares where a is 26 and b is 24. So, applying the formula, 26² - 24² would be (26 - 24)(26 + 24). Let me compute that.

26 minus 24 is 2, and 26 plus 24 is 50. So multiplying those together, 2 * 50 equals 100. So, 26² - 24² is 100. Then, the original expression inside the first parentheses is 100 minus 10. So that would be 100 - 10 = 90.

So now the entire expression simplifies to (90)² - 10². Wait, that's another difference of squares! So, (90)² - (10)² can be written as (90 - 10)(90 + 10). Let me compute that.

90 minus 10 is 80, and 90 plus 10 is 100. Multiplying those together, 80 * 100. That's straightforward: 80 * 100 = 8000.

Wait, so is the answer 8000? Let me double-check my steps to make sure I didn't make a mistake.

First step: 26² - 24². Using the difference of squares formula: (26 - 24)(26 + 24) = 2 * 50 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90². Then subtract 10². So, 90² - 10². Which is (90 - 10)(90 + 10) = 80 * 100 = 8000.

Hmm, seems correct. Let me verify by calculating each part numerically without using the formulas, just to cross-check.

Calculating 26²: 26*26. Let me compute that. 25² is 625, so 26² is 625 + 25 + 26 = 625 + 51 = 676? Wait, no. Wait, 25² is 625, then 26² is (25 + 1)² = 25² + 2*25*1 + 1² = 625 + 50 + 1 = 676. Yes, that's right.

Then 24²: 24*24. Let me compute that. 20² is 400, 4² is 16, and then cross terms: 2*20*4 = 160. So (20 + 4)² = 400 + 160 + 16 = 576. So 24² is 576.

So 26² - 24² is 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90². 90*90 is 8100. Then subtract 10², which is 100. So 8100 - 100 = 8000.

Okay, same result. So that seems to confirm that the answer is indeed 8000.

Alternatively, maybe I can think of the entire original expression as a difference of squares. Let me see. The original expression is (A)² - (10)² where A is (26² - 24² - 10). But we already found A is 90, so that's the same as before.

Alternatively, maybe there's another way to approach it? Let me try expanding everything step by step without using the difference of squares.

First, compute 26²: 676. Then 24²: 576. Then subtract them: 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² = 8100. Then subtract 10²: 8100 - 100 = 8000. Yep, same answer.

Alternatively, if I didn't notice the difference of squares, maybe I could have expanded the entire expression. Let me try that.

Original expression: (26² - 24² - 10)² - 10². Let me denote the first part as (X - 10)² - 10² where X is 26² - 24². Wait, but that's similar to before.

Alternatively, expanding (X - 10)² - 10². Let me expand (X - 10)²: X² - 20X + 100. Then subtract 10²: X² - 20X + 100 - 100 = X² - 20X. So the entire expression simplifies to X² - 20X where X is 26² - 24².

Earlier, we found X is 100. So substituting, that would be 100² - 20*100 = 10000 - 2000 = 8000. Same answer again.

So regardless of the method, I get 8000. Therefore, I think that's the correct answer.

But just to make sure, let me check if I made any miscalculations in the steps.

First, 26² is 676, 24² is 576. 676 - 576 is indeed 100. Then 100 - 10 is 90. Squared is 8100. Then subtract 10² which is 100. 8100 - 100 is 8000.

Alternatively, if I compute 90² - 10² directly: 90² is 8100, 10² is 100, difference is 8000.

Alternatively, if I didn't use the difference of squares for the first part, but just calculated 26² - 24² - 10. Let me compute 26² - 24² first. 26² is 676, 24² is 576, so 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² = 8100. Then subtract 10²: 8100 - 100 = 8000.

All methods lead to the same answer. Therefore, I can be confident that the value of the expression is 8000.

I think that's thorough enough. I considered multiple approaches, checked my calculations step by step, and verified using different methods. All lead to the same result. So I don't see any mistakes here.

**Final Answer**
\boxed{8000}

To evaluate the expression:

$$
(26^2 - 24^2 - 10)^2 - 10^2
$$

we proceed step by step, simplifying the inner parts first and applying algebraic identities where helpful.

---

### Step 1: Simplify the inner expression

We start with the inner part:

$$
26^2 - 24^2
$$

This is a **difference of squares**, which can be factored as:

$$
26^2 - 24^2 = (26 - 24)(26 + 24) = (2)(50) = 100
$$

Now subtract 10 from this result:

$$
100 - 10 = 90
$$

So the expression becomes:

$$
(90)^2 - 10^2
$$

---

### Step 2: Apply the difference of squares again

Now we have:

$$
(90)^2 - (10)^2
$$

This is another **difference of squares**, which can be factored as:

$$
(90 - 10)(90 + 10) = (80)(100) = 8000
$$

---

### Final Answer

$$
\boxed{8000}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of this expression: (26² - 24² - 10)² - 10². Hmm, let me break this down step by step. First, maybe I should compute the parts inside the parentheses first. Let me recall some algebraic identities that might help simplify things.

Wait, there's a difference of squares formula: a² - b² = (a - b)(a + b). Maybe I can use that here. Let me check if that applies to 26² - 24². Yes, that's exactly a difference of squares where a is 26 and b is 24. So, applying the formula, 26² - 24² would be (26 - 24)(26 + 24). Let me compute that.

26 minus 24 is 2, and 26 plus 24 is 50. So multiplying those together, 2 * 50 equals 100. So, 26² - 24² is 100. Then, the original expression inside the first parentheses is 100 minus 10. So that would be 100 - 10 = 90.

So now the entire expression simplifies to (90)² - 10². Wait, that's another difference of squares! So, (90)² - (10)² can be written as (90 - 10)(90 + 10). Let me compute that.

90 minus 10 is 80, and 90 plus 10 is 100. Multiplying those together, 80 * 100. That's straightforward: 80 * 100 = 8000.

Wait, so is the answer 8000? Let me double-check my steps to make sure I didn't make a mistake.

First step: 26² - 24². Using the difference of squares formula: (26 - 24)(26 + 24) = 2 * 50 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90². Then subtract 10². So, 90² - 10². Which is (90 - 10)(90 + 10) = 80 * 100 = 8000.

Hmm, seems correct. Let me verify by calculating each part numerically without using the formulas, just to cross-check.

Calculating 26²: 26*26. Let me compute that. 25² is 625, so 26² is 625 + 25 + 26 = 625 + 51 = 676? Wait, no. Wait, 25² is 625, then 26² is (25 + 1)² = 25² + 2*25*1 + 1² = 625 + 50 + 1 = 676. Yes, that's right.

Then 24²: 24*24. Let me compute that. 20² is 400, 4² is 16, and then cross terms: 2*20*4 = 160. So (20 + 4)² = 400 + 160 + 16 = 576. So 24² is 576.

So 26² - 24² is 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90². 90*90 is 8100. Then subtract 10², which is 100. So 8100 - 100 = 8000.

Okay, same result. So that seems to confirm that the answer is indeed 8000.

Alternatively, maybe I can think of the entire original expression as a difference of squares. Let me see. The original expression is (A)² - (10)² where A is (26² - 24² - 10). But we already found A is 90, so that's the same as before.

Alternatively, maybe there's another way to approach it? Let me try expanding everything step by step without using the difference of squares.

First, compute 26²: 676. Then 24²: 576. Then subtract them: 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² = 8100. Then subtract 10²: 8100 - 100 = 8000. Yep, same answer.

Alternatively, if I didn't notice the difference of squares, maybe I could have expanded the entire expression. Let me try that.

Original expression: (26² - 24² - 10)² - 10². Let me denote the first part as (X - 10)² - 10² where X is 26² - 24². Wait, but that's similar to before.

Alternatively, expanding (X - 10)² - 10². Let me expand (X - 10)²: X² - 20X + 100. Then subtract 10²: X² - 20X + 100 - 100 = X² - 20X. So the entire expression simplifies to X² - 20X where X is 26² - 24².

Earlier, we found X is 100. So substituting, that would be 100² - 20*100 = 10000 - 2000 = 8000. Same answer again.

So regardless of the method, I get 8000. Therefore, I think that's the correct answer.

But just to make sure, let me check if I made any miscalculations in the steps.

First, 26² is 676, 24² is 576. 676 - 576 is indeed 100. Then 100 - 10 is 90. Squared is 8100. Then subtract 10² which is 100. 8100 - 100 is 8000.

Alternatively, if I compute 90² - 10² directly: 90² is 8100, 10² is 100, difference is 8000.

Alternatively, if I didn't use the difference of squares for the first part, but just calculated 26² - 24² - 10. Let me compute 26² - 24² first. 26² is 676, 24² is 576, so 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² = 8100. Then subtract 10²: 8100 - 100 = 8000.

All methods lead to the same answer. Therefore, I can be confident that the value of the expression is 8000.

I think that's thorough enough. I considered multiple approaches, checked my calculations step by step, and verified using different methods. All lead to the same result. So I don't see any mistakes here.

**Final Answer**
\boxed{8000}

To evaluate the expression:

$$
(26^2 - 24^2 - 10)^2 - 10^2
$$

we proceed step by step, simplifying the inner parts first and applying algebraic identities where helpful.

---

### Step 1: Simplify the inner expression

We start with the inner part:

$$
26^2 - 24^2
$$

This is a **difference of squares**, which can be factored as:

$$
26^2 - 24^2 = (26 - 24)(26 + 24) = (2)(50) = 100
$$

Now subtract 10 from this result:

$$
100 - 10 = 90
$$

So the expression becomes:

$$
(90)^2 - 10^2
$$

---

### Step 2: Apply the difference of squares again

Now we have:

$$
(90)^2 - (10)^2
$$

This is another **difference of squares**, which can be factored as:

$$
(90 - 10)(90 + 10) = (80)(100) = 8000
$$

---

### Final Answer

$$
\boxed{8000}
$$

Solution: We know that $x^2 - y^2 = (x+y)(x-y)$. To begin with, let $x = 26^2 - 24^2 - 10$ and $y = 10$. Factoring $x^2 - y^2$ and substituting in gives $(26^2-24^2-10+10)(26^2-24^2-10-10)$. Now, let $x = 26$ and $y = 24$. Factoring $x^2 - y^2$ and substituting in gives $((26+24)(26-24)-10+10)((26+24)(26-24)-10-10)$. This simplifies to $(50\cdot 2)(50 \cdot 2 - 20)$, or $100 \cdot 80$. Thus, our final answer is $\boxed{8000}$.


Explanation:
Okay, so I need to figure out how many ways there are to arrange the letters of the word "ELLIPSE". Let me start by recalling what I know about permutations of letters with repeated characters.

First, I remember that if all the letters in a word are unique, the number of arrangements is just the factorial of the number of letters. For example, a 5-letter word with all unique letters would have 5! = 120 arrangements. But when there are repeating letters, we have to adjust for those repetitions because some arrangements would be duplicates.

So, let me check the letters in "ELLIPSE". Let me write them out: E, L, L, I, P, S, E. Wait, let me count them again. E, L, L, I, P, S, E. That's 7 letters total. Let me check for duplicates.

Looking at each letter:

- E appears twice
- L appears twice
- The other letters are I, P, S, which are unique.

So, the letters are: E, E, L, L, I, P, S. So, two E's and two L's, and the rest are single.

Therefore, the formula for permutations of multiset is n! divided by the product of (k_i!) for each repeated element, where k_i is the number of times each element is repeated.

So, in this case, n is 7 (total letters), and we have two letters that are repeated: E and L, each appearing twice. Therefore, the number of distinct arrangements should be 7! divided by (2! * 2!).

Let me verify that.

First, compute 7! which is 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me calculate that.

7! = 5040.

Then, 2! is 2, so 2! * 2! = 2 * 2 = 4.

Therefore, 5040 divided by 4 is 1260.

Wait, so is the answer 1260? Let me check again to make sure I didn't make a mistake.

Alternatively, maybe I can think of it step by step.

Suppose all letters were unique. Then there would be 7! arrangements. But since there are duplicates, we have to divide by the number of ways the duplicates can be arranged among themselves.

For example, the two E's: if they were distinct, say E1 and E2, then swapping them would result in the same word. Similarly for the two L's.

Therefore, the total number of duplicates is 2! for the E's and 2! for the L's. Therefore, dividing by 2! * 2! gives the correct count.

So, 7! / (2! * 2!) = 5040 / 4 = 1260.

But let me check if there are any other repeated letters. The letters are E, L, L, I, P, S, E. So, E is twice, L is twice, and the rest are once. So, yes, only two letters are repeated.

Alternatively, maybe I can think of it as first choosing positions for the E's, then positions for the L's, and then arranging the rest.

So, first, how many ways to arrange the letters? Let me think.

Total letters: 7.

First, choose positions for the E's. There are C(7,2) ways to choose positions for the two E's. Then, from the remaining 5 positions, choose positions for the two L's: C(5,2). Then, the remaining 3 positions are for the letters I, P, S, which are distinct, so they can be arranged in 3! ways.

Therefore, total number of arrangements is C(7,2) * C(5,2) * 3!.

Let me compute that.

C(7,2) is 21, C(5,2) is 10, and 3! is 6.

So, 21 * 10 * 6 = 21 * 60 = 1260.

Same result. So that seems to confirm the answer.

Alternatively, if I use the formula for permutations with repetition: n! divided by the product of the factorials of the counts of each repeated element.

Which is exactly what I did first.

So, both methods give me 1260.

Wait, but let me check if there's any other possible mistake here. For example, maybe I miscounted the letters? Let me check again.

The word is ELLIPSE. Let me spell it out: E, L, L, I, P, S, E. Yes, that's 7 letters.

E appears twice, L appears twice, and I, P, S once each. So, yes, two pairs of duplicates.

Therefore, the formula applies.

Alternatively, maybe I can think of the total permutations as 7! and then divide by the duplicates.

But why divide by 2! for E and 2! for L? Because for each set of duplicates, the number of ways they can be arranged among themselves is 2! for E and 2! for L, so we divide by those to account for overcounting.

Yes, that's the standard approach.

Alternatively, if I think of the total number of permutations as 7! and then divide by the number of duplicates.

But in this case, since there are two different letters that are duplicated, each with two copies, so dividing by 2! for each.

So, 7! / (2! * 2!) = 5040 / 4 = 1260.

Therefore, I think that is correct.

But just to be thorough, let me try a smaller example to see if my reasoning holds.

Suppose I have the word "AABB". How many arrangements?

Using the formula: 4! / (2! * 2!) = 24 / 4 = 6.

Listing them out: AABB, ABAB, ABBA, BAAB, BABA, BBAA. Yes, that's 6.

Alternatively, using the other method: choose positions for A's: C(4,2) = 6, then the B's are in the remaining positions. So, same result.

So that works.

Another example: "AABC". Letters A, A, B, C. Number of arrangements: 4! / 2! = 24 / 2 = 12.

Alternatively, choosing positions for A's: C(4,2) = 6, then arranging B and C in the remaining 2 positions: 2! = 2. So total 6 * 2 = 12. Correct.

So, same approach.

Therefore, applying that to "ELLIPSE", which has two pairs of duplicates, the formula is 7! / (2! * 2!) = 1260.

Therefore, I think that is the correct answer.

But just to make sure, let me think if there's any other possible mistake. For example, maybe the word has more repeated letters? Let me check again.

ELLIPSE: E, L, L, I, P, S, E.

Letters: E (twice), L (twice), I (once), P (once), S (once).

No, that's correct. So, only two letters are repeated.

Alternatively, maybe I misread the word? Let me check again.

ELLIPSE. E-L-L-I-P-S-E. Yes, that's correct.

So, the letters are E, L, L, I, P, S, E.

Yes, so two E's and two L's.

Therefore, the calculation is correct.

Alternatively, if I think of the letters as E, E, L, L, I, P, S.

So, total letters: 7.

Number of distinct arrangements: 7! divided by (2! for E's and 2! for L's).

Yes.

Alternatively, if I think of the problem as first arranging all letters, considering duplicates.

So, if all letters were unique, it's 7! But since two letters are duplicated, we have to divide by the number of ways those duplicates can be arranged.

For example, if we have two E's, then swapping them doesn't create a new arrangement, so we divide by 2! for the E's. Similarly for the L's.

Therefore, total number is 7! / (2! * 2!) = 1260.

Therefore, I think that is correct.

Alternatively, maybe I can think of the problem as follows:

Suppose I have 7 positions.

First, choose positions for the E's: C(7,2).

Then, choose positions for the L's: C(5,2).

Then, the remaining 3 positions are for I, P, S. Since they are distinct, they can be arranged in 3! ways.

So, total number is C(7,2) * C(5,2) * 3!

Calculating:

C(7,2) is 21, C(5,2) is 10, 3! is 6.

21 * 10 = 210, 210 * 6 = 1260.

Same answer.

Therefore, this method also gives 1260.

Therefore, I think that is correct.

Alternatively, if I use the multinomial coefficient.

The multinomial coefficient for dividing n items into groups of sizes k1, k2, ..., km is n! / (k1! * k2! * ... * km!).

In this case, we have groups: two E's, two L's, one I, one P, one S.

Therefore, the number of arrangements is 7! / (2! * 2! * 1! * 1! * 1!) = 7! / (2! * 2!) = same as before.

Yes, because the 1! terms don't affect the calculation.

Therefore, again, 5040 / 4 = 1260.

Therefore, all methods lead to the same answer.

Therefore, I think the answer is 1260.

But just to be 100% sure, let me check with another approach.

Suppose I try to calculate the total permutations and then subtract the duplicates. Wait, but that might be more complicated.

Alternatively, think of it as permutations with identical elements.

Yes, which is exactly what we did.

Alternatively, think of generating all permutations and then dividing by the duplicates.

But since the formula is standard, and multiple methods confirm the same answer, I think it's safe.

Therefore, I think the number of distinct arrangements of the letters in "ELLIPSE" is 1260.

**Final Answer**
The number of ways to arrange the letters of the word ELLIPSE is \boxed{1260}.

To determine the number of distinct ways to arrange the letters of the word **"ELLIPSE"**, we begin by analyzing the composition of the word.

---

### Step 1: Count the Total Letters and Identify Repeats

The word **"ELLIPSE"** consists of the following letters:

- **E** appears **2 times**
- **L** appears **2 times**
- **I**, **P**, and **S** each appear **1 time**

So, the total number of letters is:

$$
n = 7
$$

And the repeated letters are:

- E: 2 times
- L: 2 times

---

### Step 2: Apply the Permutations with Repeated Elements Formula

When a word contains repeated letters, the number of **distinct permutations** is given by the formula:

$$
\text{Number of arrangements} = \frac{n!}{k_1! \cdot k_2! \cdot \ldots \cdot k_m!}
$$

Where:
- $ n $ is the total number of letters
- $ k_1, k_2, \ldots, k_m $ are the frequencies of each repeated letter

In this case:

$$
\text{Number of arrangements} = \frac{7!}{2! \cdot 2!}
$$

---

### Step 3: Compute the Values

- $ 7! = 5040 $
- $ 2! = 2 $, so $ 2! \cdot 2! = 2 \cdot 2 = 4 $

$$
\text{Number of arrangements} = \frac{5040}{4} = 1260
$$

---

### Step 4: Confirm with an Alternative Method

We can also compute this by first choosing positions for the repeated letters:

- Choose 2 positions out of 7 for the E's: $ \binom{7}{2} = 21 $
- From the remaining 5 positions, choose 2 for the L's: $ \binom{5}{2} = 10 $
- The remaining 3 positions are for the distinct letters I, P, and S, which can be arranged in $ 3! = 6 $ ways

So the total number of arrangements is:

$$
\binom{7}{2} \cdot \binom{5}{2} \cdot 3! = 21 \cdot 10 \cdot 6 = 1260
$$

This confirms the result.

---

### Final Answer

$$
\boxed{1260}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways there are to arrange the letters of the word "ELLIPSE". Let me start by recalling what I know about permutations of letters with repeated characters.

First, I remember that if all the letters in a word are unique, the number of arrangements is just the factorial of the number of letters. For example, a 5-letter word with all unique letters would have 5! = 120 arrangements. But when there are repeating letters, we have to adjust for those repetitions because some arrangements would be duplicates.

So, let me check the letters in "ELLIPSE". Let me write them out: E, L, L, I, P, S, E. Wait, let me count them again. E, L, L, I, P, S, E. That's 7 letters total. Let me check for duplicates.

Looking at each letter:

- E appears twice
- L appears twice
- The other letters are I, P, S, which are unique.

So, the letters are: E, E, L, L, I, P, S. So, two E's and two L's, and the rest are single.

Therefore, the formula for permutations of multiset is n! divided by the product of (k_i!) for each repeated element, where k_i is the number of times each element is repeated.

So, in this case, n is 7 (total letters), and we have two letters that are repeated: E and L, each appearing twice. Therefore, the number of distinct arrangements should be 7! divided by (2! * 2!).

Let me verify that.

First, compute 7! which is 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me calculate that.

7! = 5040.

Then, 2! is 2, so 2! * 2! = 2 * 2 = 4.

Therefore, 5040 divided by 4 is 1260.

Wait, so is the answer 1260? Let me check again to make sure I didn't make a mistake.

Alternatively, maybe I can think of it step by step.

Suppose all letters were unique. Then there would be 7! arrangements. But since there are duplicates, we have to divide by the number of ways the duplicates can be arranged among themselves.

For example, the two E's: if they were distinct, say E1 and E2, then swapping them would result in the same word. Similarly for the two L's.

Therefore, the total number of duplicates is 2! for the E's and 2! for the L's. Therefore, dividing by 2! * 2! gives the correct count.

So, 7! / (2! * 2!) = 5040 / 4 = 1260.

But let me check if there are any other repeated letters. The letters are E, L, L, I, P, S, E. So, E is twice, L is twice, and the rest are once. So, yes, only two letters are repeated.

Alternatively, maybe I can think of it as first choosing positions for the E's, then positions for the L's, and then arranging the rest.

So, first, how many ways to arrange the letters? Let me think.

Total letters: 7.

First, choose positions for the E's. There are C(7,2) ways to choose positions for the two E's. Then, from the remaining 5 positions, choose positions for the two L's: C(5,2). Then, the remaining 3 positions are for the letters I, P, S, which are distinct, so they can be arranged in 3! ways.

Therefore, total number of arrangements is C(7,2) * C(5,2) * 3!.

Let me compute that.

C(7,2) is 21, C(5,2) is 10, and 3! is 6.

So, 21 * 10 * 6 = 21 * 60 = 1260.

Same result. So that seems to confirm the answer.

Alternatively, if I use the formula for permutations with repetition: n! divided by the product of the factorials of the counts of each repeated element.

Which is exactly what I did first.

So, both methods give me 1260.

Wait, but let me check if there's any other possible mistake here. For example, maybe I miscounted the letters? Let me check again.

The word is ELLIPSE. Let me spell it out: E, L, L, I, P, S, E. Yes, that's 7 letters.

E appears twice, L appears twice, and I, P, S once each. So, yes, two pairs of duplicates.

Therefore, the formula applies.

Alternatively, maybe I can think of the total permutations as 7! and then divide by the duplicates.

But why divide by 2! for E and 2! for L? Because for each set of duplicates, the number of ways they can be arranged among themselves is 2! for E and 2! for L, so we divide by those to account for overcounting.

Yes, that's the standard approach.

Alternatively, if I think of the total number of permutations as 7! and then divide by the number of duplicates.

But in this case, since there are two different letters that are duplicated, each with two copies, so dividing by 2! for each.

So, 7! / (2! * 2!) = 5040 / 4 = 1260.

Therefore, I think that is correct.

But just to be thorough, let me try a smaller example to see if my reasoning holds.

Suppose I have the word "AABB". How many arrangements?

Using the formula: 4! / (2! * 2!) = 24 / 4 = 6.

Listing them out: AABB, ABAB, ABBA, BAAB, BABA, BBAA. Yes, that's 6.

Alternatively, using the other method: choose positions for A's: C(4,2) = 6, then the B's are in the remaining positions. So, same result.

So that works.

Another example: "AABC". Letters A, A, B, C. Number of arrangements: 4! / 2! = 24 / 2 = 12.

Alternatively, choosing positions for A's: C(4,2) = 6, then arranging B and C in the remaining 2 positions: 2! = 2. So total 6 * 2 = 12. Correct.

So, same approach.

Therefore, applying that to "ELLIPSE", which has two pairs of duplicates, the formula is 7! / (2! * 2!) = 1260.

Therefore, I think that is the correct answer.

But just to make sure, let me think if there's any other possible mistake. For example, maybe the word has more repeated letters? Let me check again.

ELLIPSE: E, L, L, I, P, S, E.

Letters: E (twice), L (twice), I (once), P (once), S (once).

No, that's correct. So, only two letters are repeated.

Alternatively, maybe I misread the word? Let me check again.

ELLIPSE. E-L-L-I-P-S-E. Yes, that's correct.

So, the letters are E, L, L, I, P, S, E.

Yes, so two E's and two L's.

Therefore, the calculation is correct.

Alternatively, if I think of the letters as E, E, L, L, I, P, S.

So, total letters: 7.

Number of distinct arrangements: 7! divided by (2! for E's and 2! for L's).

Yes.

Alternatively, if I think of the problem as first arranging all letters, considering duplicates.

So, if all letters were unique, it's 7! But since two letters are duplicated, we have to divide by the number of ways those duplicates can be arranged.

For example, if we have two E's, then swapping them doesn't create a new arrangement, so we divide by 2! for the E's. Similarly for the L's.

Therefore, total number is 7! / (2! * 2!) = 1260.

Therefore, I think that is correct.

Alternatively, maybe I can think of the problem as follows:

Suppose I have 7 positions.

First, choose positions for the E's: C(7,2).

Then, choose positions for the L's: C(5,2).

Then, the remaining 3 positions are for I, P, S. Since they are distinct, they can be arranged in 3! ways.

So, total number is C(7,2) * C(5,2) * 3!

Calculating:

C(7,2) is 21, C(5,2) is 10, 3! is 6.

21 * 10 = 210, 210 * 6 = 1260.

Same answer.

Therefore, this method also gives 1260.

Therefore, I think that is correct.

Alternatively, if I use the multinomial coefficient.

The multinomial coefficient for dividing n items into groups of sizes k1, k2, ..., km is n! / (k1! * k2! * ... * km!).

In this case, we have groups: two E's, two L's, one I, one P, one S.

Therefore, the number of arrangements is 7! / (2! * 2! * 1! * 1! * 1!) = 7! / (2! * 2!) = same as before.

Yes, because the 1! terms don't affect the calculation.

Therefore, again, 5040 / 4 = 1260.

Therefore, all methods lead to the same answer.

Therefore, I think the answer is 1260.

But just to be 100% sure, let me check with another approach.

Suppose I try to calculate the total permutations and then subtract the duplicates. Wait, but that might be more complicated.

Alternatively, think of it as permutations with identical elements.

Yes, which is exactly what we did.

Alternatively, think of generating all permutations and then dividing by the duplicates.

But since the formula is standard, and multiple methods confirm the same answer, I think it's safe.

Therefore, I think the number of distinct arrangements of the letters in "ELLIPSE" is 1260.

**Final Answer**
The number of ways to arrange the letters of the word ELLIPSE is \boxed{1260}.

To determine the number of distinct ways to arrange the letters of the word **"ELLIPSE"**, we begin by analyzing the composition of the word.

---

### Step 1: Count the Total Letters and Identify Repeats

The word **"ELLIPSE"** consists of the following letters:

- **E** appears **2 times**
- **L** appears **2 times**
- **I**, **P**, and **S** each appear **1 time**

So, the total number of letters is:

$$
n = 7
$$

And the repeated letters are:

- E: 2 times
- L: 2 times

---

### Step 2: Apply the Permutations with Repeated Elements Formula

When a word contains repeated letters, the number of **distinct permutations** is given by the formula:

$$
\text{Number of arrangements} = \frac{n!}{k_1! \cdot k_2! \cdot \ldots \cdot k_m!}
$$

Where:
- $ n $ is the total number of letters
- $ k_1, k_2, \ldots, k_m $ are the frequencies of each repeated letter

In this case:

$$
\text{Number of arrangements} = \frac{7!}{2! \cdot 2!}
$$

---

### Step 3: Compute the Values

- $ 7! = 5040 $
- $ 2! = 2 $, so $ 2! \cdot 2! = 2 \cdot 2 = 4 $

$$
\text{Number of arrangements} = \frac{5040}{4} = 1260
$$

---

### Step 4: Confirm with an Alternative Method

We can also compute this by first choosing positions for the repeated letters:

- Choose 2 positions out of 7 for the E's: $ \binom{7}{2} = 21 $
- From the remaining 5 positions, choose 2 for the L's: $ \binom{5}{2} = 10 $
- The remaining 3 positions are for the distinct letters I, P, and S, which can be arranged in $ 3! = 6 $ ways

So the total number of arrangements is:

$$
\binom{7}{2} \cdot \binom{5}{2} \cdot 3! = 21 \cdot 10 \cdot 6 = 1260
$$

This confirms the result.

---

### Final Answer

$$
\boxed{1260}
$$

Solution: There are two E's, two L's, and seven total letters, so the answer is $\dfrac{7!}{2! \times 2!} = \boxed{1260}$.


Explanation:
Okay, so I need to solve the equation $2^{2x} = 256^{\frac{1}{2}}$. Hmm, let me think about how to approach this.

First, I remember that when dealing with exponential equations, it's often helpful to express both sides with the same base if possible. That way, I can set the exponents equal to each other. Let me check if 256 can be written as a power of 2.

I know that $2^8 = 256$ because $2^10 is 1024, so 2^8 is 256. Let me verify that: 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, 2^6 is 64, 2^7 is 128, and 2^8 is 256. Yep, that's right. So 256 is $2^8$.

So substituting that into the original equation, the right side becomes $(2^8)^{\frac{1}{2}}$. Now, I need to simplify that exponent. Remember that when you raise a power to another power, you multiply the exponents. So $(2^8)^{\frac{1}{2}} = 2^{8 \times \frac{1}{2}} = 2^4$.

So now the equation is $2^{2x} = 2^4$. Since the bases are the same (both are base 2), I can set the exponents equal to each other. That gives me $2x = 4$.

Solving for x, I just need to divide both sides by 2. So $x = 4 / 2 = 2$.

Wait, let me double-check my steps to make sure I didn't make a mistake. Starting with the original equation: $2^{2x} = 256^{1/2}$.

Expressed as powers of 2, 256 is indeed 2^8, so the right side is (2^8)^{1/2} which simplifies to 2^{4}. Then, 2^{2x} equals 2^4, so exponents must be equal: 2x = 4, so x = 2.

Let me plug the value back into the original equation to verify.

Left side: $2^{2x}$ when x=2 is $2^{2*2} = 2^4 = 16$.

Right side: $256^{1/2}$ is the square root of 256, which is 16.

Yes, both sides equal 16, so x=2 is correct.

Alternatively, maybe I can approach this problem another way to confirm. Let me try taking the logarithm of both sides.

Taking natural logarithm on both sides:

$\ln(2^{2x}) = \ln(256^{1/2})$

Using the logarithm power rule, which says $\ln(a^b) = b\ln(a)$, so:

$2x \ln 2 = \frac{1}{2} \ln 256$

Then, solving for x:

$x = \frac{\frac{1}{2} \ln 256}{2 \ln 2} = \frac{\ln 256}{4 \ln 2}$

But since 256 is 2^8, $\ln 256 = \ln(2^8) = 8 \ln 2$. Substituting that in:

$x = \frac{8 \ln 2}{4 \ln 2} = \frac{8}{4} = 2$.

Same result. So that confirms it again.

Another way: Maybe express both sides as exponents with base 16 or something else? Let me see.

But 256 is 16 squared, since 16^2 is 256. So 256^{1/2} is (16^2)^{1/2} = 16^{1} = 16. Then, original equation is 2^{2x} = 16.

But 16 is 2^4, so again 2^{2x} = 2^4, so 2x=4, x=2.

Same answer.

Alternatively, maybe I can think about the original equation as (2^2)^x = 256^{1/2}.

Since 2^2 is 4, so 4^x = 256^{1/2}.

Then, 256^{1/2} is 16, so 4^x = 16.

But 16 is 4^2, so x=2.

Same result again.

So regardless of the method, I keep getting x=2.

I think that's solid. Maybe I can check with decimal exponents? Let me see.

Original equation: 2^{2x} = sqrt(256).

sqrt(256) is 16. So 2^{2x} = 16.

If I take log base 2 of both sides: log2(2^{2x}) = log2(16).

Left side simplifies to 2x. Right side is log2(16). Since 16 is 2^4, log2(16)=4.

So 2x=4, x=2.

Same answer.

I think I've verified it multiple ways. So I'm confident that x=2 is the correct solution.

Just to recap all steps clearly:

1. Recognize that 256 is a power of 2 (specifically, 2^8).
2. Rewrite the equation with the same base: 2^{2x} = (2^8)^{1/2}.
3. Simplify the right side using exponent rules: 2^{8*(1/2)} = 2^4.
4. Now the equation is 2^{2x} = 2^4, so set exponents equal: 2x = 4.
5. Solve for x: x = 2.
6. Verify by plugging back into original equation: both sides equal 16, confirming the solution.

No steps seem to have errors, and multiple methods lead to the same answer. Therefore, the solution is correct.

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

To solve the equation:

$$
2^{2x} = 256^{\frac{1}{2}}
$$

we begin by simplifying both sides of the equation using exponent rules and properties of powers.

---

### Step 1: Express 256 as a power of 2

We know that:

$$
256 = 2^8
$$

So, we can rewrite the right-hand side of the equation as:

$$
256^{\frac{1}{2}} = (2^8)^{\frac{1}{2}}
$$

---

### Step 2: Apply the power rule $(a^m)^n = a^{m \cdot n}$

$$
(2^8)^{\frac{1}{2}} = 2^{8 \cdot \frac{1}{2}} = 2^4
$$

Now the equation becomes:

$$
2^{2x} = 2^4
$$

---

### Step 3: Set the exponents equal

Since the bases are the same (both are base 2), we can equate the exponents:

$$
2x = 4
$$

---

### Step 4: Solve for $x$

$$
x = \frac{4}{2} = 2
$$

---

### Step 5: Verify the solution

Substitute $x = 2$ back into the original equation:

- Left-hand side: $2^{2x} = 2^{2 \cdot 2} = 2^4 = 16$
- Right-hand side: $256^{\frac{1}{2}} = \sqrt{256} = 16$

Both sides are equal, confirming the solution is correct.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation $2^{2x} = 256^{\frac{1}{2}}$. Hmm, let me think about how to approach this.

First, I remember that when dealing with exponential equations, it's often helpful to express both sides with the same base if possible. That way, I can set the exponents equal to each other. Let me check if 256 can be written as a power of 2.

I know that $2^8 = 256$ because $2^10 is 1024, so 2^8 is 256. Let me verify that: 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, 2^6 is 64, 2^7 is 128, and 2^8 is 256. Yep, that's right. So 256 is $2^8$.

So substituting that into the original equation, the right side becomes $(2^8)^{\frac{1}{2}}$. Now, I need to simplify that exponent. Remember that when you raise a power to another power, you multiply the exponents. So $(2^8)^{\frac{1}{2}} = 2^{8 \times \frac{1}{2}} = 2^4$.

So now the equation is $2^{2x} = 2^4$. Since the bases are the same (both are base 2), I can set the exponents equal to each other. That gives me $2x = 4$.

Solving for x, I just need to divide both sides by 2. So $x = 4 / 2 = 2$.

Wait, let me double-check my steps to make sure I didn't make a mistake. Starting with the original equation: $2^{2x} = 256^{1/2}$.

Expressed as powers of 2, 256 is indeed 2^8, so the right side is (2^8)^{1/2} which simplifies to 2^{4}. Then, 2^{2x} equals 2^4, so exponents must be equal: 2x = 4, so x = 2.

Let me plug the value back into the original equation to verify.

Left side: $2^{2x}$ when x=2 is $2^{2*2} = 2^4 = 16$.

Right side: $256^{1/2}$ is the square root of 256, which is 16.

Yes, both sides equal 16, so x=2 is correct.

Alternatively, maybe I can approach this problem another way to confirm. Let me try taking the logarithm of both sides.

Taking natural logarithm on both sides:

$\ln(2^{2x}) = \ln(256^{1/2})$

Using the logarithm power rule, which says $\ln(a^b) = b\ln(a)$, so:

$2x \ln 2 = \frac{1}{2} \ln 256$

Then, solving for x:

$x = \frac{\frac{1}{2} \ln 256}{2 \ln 2} = \frac{\ln 256}{4 \ln 2}$

But since 256 is 2^8, $\ln 256 = \ln(2^8) = 8 \ln 2$. Substituting that in:

$x = \frac{8 \ln 2}{4 \ln 2} = \frac{8}{4} = 2$.

Same result. So that confirms it again.

Another way: Maybe express both sides as exponents with base 16 or something else? Let me see.

But 256 is 16 squared, since 16^2 is 256. So 256^{1/2} is (16^2)^{1/2} = 16^{1} = 16. Then, original equation is 2^{2x} = 16.

But 16 is 2^4, so again 2^{2x} = 2^4, so 2x=4, x=2.

Same answer.

Alternatively, maybe I can think about the original equation as (2^2)^x = 256^{1/2}.

Since 2^2 is 4, so 4^x = 256^{1/2}.

Then, 256^{1/2} is 16, so 4^x = 16.

But 16 is 4^2, so x=2.

Same result again.

So regardless of the method, I keep getting x=2.

I think that's solid. Maybe I can check with decimal exponents? Let me see.

Original equation: 2^{2x} = sqrt(256).

sqrt(256) is 16. So 2^{2x} = 16.

If I take log base 2 of both sides: log2(2^{2x}) = log2(16).

Left side simplifies to 2x. Right side is log2(16). Since 16 is 2^4, log2(16)=4.

So 2x=4, x=2.

Same answer.

I think I've verified it multiple ways. So I'm confident that x=2 is the correct solution.

Just to recap all steps clearly:

1. Recognize that 256 is a power of 2 (specifically, 2^8).
2. Rewrite the equation with the same base: 2^{2x} = (2^8)^{1/2}.
3. Simplify the right side using exponent rules: 2^{8*(1/2)} = 2^4.
4. Now the equation is 2^{2x} = 2^4, so set exponents equal: 2x = 4.
5. Solve for x: x = 2.
6. Verify by plugging back into original equation: both sides equal 16, confirming the solution.

No steps seem to have errors, and multiple methods lead to the same answer. Therefore, the solution is correct.

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

To solve the equation:

$$
2^{2x} = 256^{\frac{1}{2}}
$$

we begin by simplifying both sides of the equation using exponent rules and properties of powers.

---

### Step 1: Express 256 as a power of 2

We know that:

$$
256 = 2^8
$$

So, we can rewrite the right-hand side of the equation as:

$$
256^{\frac{1}{2}} = (2^8)^{\frac{1}{2}}
$$

---

### Step 2: Apply the power rule $(a^m)^n = a^{m \cdot n}$

$$
(2^8)^{\frac{1}{2}} = 2^{8 \cdot \frac{1}{2}} = 2^4
$$

Now the equation becomes:

$$
2^{2x} = 2^4
$$

---

### Step 3: Set the exponents equal

Since the bases are the same (both are base 2), we can equate the exponents:

$$
2x = 4
$$

---

### Step 4: Solve for $x$

$$
x = \frac{4}{2} = 2
$$

---

### Step 5: Verify the solution

Substitute $x = 2$ back into the original equation:

- Left-hand side: $2^{2x} = 2^{2 \cdot 2} = 2^4 = 16$
- Right-hand side: $256^{\frac{1}{2}} = \sqrt{256} = 16$

Both sides are equal, confirming the solution is correct.

---

### Final Answer

$$
\boxed{2}
$$

Solution:
\begin{align*}
2^{2x} & =256^{\frac{1}{2}} \\
2^{2x} & =(2^8)^{\frac{1}{2}} \\
2^{2x} & =(2^4) \\
2x & = 4 \\
x & = \boxed{2}
\end{align*}