Select the correct answer to the following equation: 27x^3/4−1=26 .(1 point)

To solve the equation \( 27x^{3/4} - 1 = 26 \), we first isolate the term involving \( x \):

1. Add 1 to both sides:
\[
27x^{3/4} = 27
\]

2. Divide both sides by 27:
\[
x^{3/4} = 1
\]

3. To eliminate the exponent \( \frac{3}{4} \), raise both sides to the reciprocal power, \( \frac{4}{3} \):
\[
x = 1^{\frac{4}{3}} = 1
\]

So the solution to the equation is \( x = 1 \).

The correct answer is:

**1**

The volume of an open top box is created by cutting a square from each corner of a 10 in. by 12 in. piece of paper. The situation is represented by the function V(x)=x(10−2x)(12−2x)

. Find the maximum volume of a box that can be created from the piece of paper. Round the answer to one decimal place.

To find the maximum volume of the open-top box formed by cutting a square of side \( x \) from each corner of a \( 10 \) inch by \( 12 \) inch piece of paper, we use the volume function given by:

\[
V(x) = x(10 - 2x)(12 - 2x)
\]

First, we simplify the expression for \( V(x) \):

1. Expand \( V(x) \):
\[
V(x) = x(10 - 2x)(12 - 2x)
\]

2. We can expand \( (10 - 2x)(12 - 2x) \) first:
\[
(10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2
\]

3. Now substitute this back into the expression for \( V(x) \):
\[
V(x) = x(120 - 44x + 4x^2)
\]
\[
V(x) = 120x - 44x^2 + 4x^3
\]

To find the maximum volume, we take the derivative of \( V(x) \) and set it to zero:

4. Find \( V'(x) \):
\[
V'(x) = 120 - 88x + 12x^2
\]

5. Set the derivative equal to zero to find the critical points:
\[
12x^2 - 88x + 120 = 0
\]

6. Simplify the equation by dividing by 4:
\[
3x^2 - 22x + 30 = 0
\]

7. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3}
\]
\[
= \frac{22 \pm \sqrt{484 - 360}}{6}
\]
\[
= \frac{22 \pm \sqrt{124}}{6}
\]
\[
= \frac{22 \pm 2\sqrt{31}}{6}
\]
\[
= \frac{11 \pm \sqrt{31}}{3}
\]

8. Evaluate the approximate solutions:
- For \( x = \frac{11 + \sqrt{31}}{3} \approx 5.27 \) (not feasible since \( 5.27 > 5 \))
- For \( x = \frac{11 - \sqrt{31}}{3} \approx 1.73 \)

Now, we substitute \( x = 1.73 \) back into the volume function to find the maximum volume:

9. Calculate \( V(1.73) \):
\[
V(1.73) = 1.73(10 - 2 \cdot 1.73)(12 - 2 \cdot 1.73)
\]
\[
= 1.73(10 - 3.46)(12 - 3.46)
\]
\[
= 1.73(6.54)(8.54)
\]
\[
= 1.73 \cdot 55.8976 \approx 96.56
\]

Thus, the maximum volume of the box that can be created is approximately \( 96.6 \) cubic inches when rounded to one decimal place.

The final answer is:

**96.6 cubic inches**.

no. it's around 96.8

Let’s reanalyze the problem and recalculate the volume to ensure accuracy.

We need to find the maximum volume of the open-top box defined by the volume function:

\[
V(x) = x(10 - 2x)(12 - 2x).
\]

### Step 1: Expand and Simplify

First, we'll simplify the volume function:

1. **Expand the terms:**
- \((10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2.\)

2. **Thus, we have:**
\[
V(x) = x(120 - 44x + 4x^2) = 120x - 44x^2 + 4x^3.
\]

### Step 2: Find the Derivative

Next, find the derivative of \( V(x) \):
\[
V'(x) = 120 - 88x + 12x^2.
\]

### Step 3: Set the Derivative to Zero

To find critical points, set the derivative equal to zero:

\[
12x^2 - 88x + 120 = 0.
\]

### Step 4: Use the Quadratic Formula

Divide by 4:
\[
3x^2 - 22x + 30 = 0.
\]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

- \( a = 3 \)
- \( b = -22 \)
- \( c = 30 \)

\[
x = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3}
= \frac{22 \pm \sqrt{484 - 360}}{6}
= \frac{22 \pm \sqrt{124}}{6}
= \frac{22 \pm 2\sqrt{31}}{6}
= \frac{11 \pm \sqrt{31}}{3}.
\]

### Step 5: Evaluate the Critical Points

Calculating the discriminant:
- \(\sqrt{31} \approx 5.57\).

So:
\[
x_1 = \frac{11 + \sqrt{31}}{3} \approx \frac{11 + 5.57}{3} \approx 5.19 \quad (\text{not feasible}).
\]
\[
x_2 = \frac{11 - \sqrt{31}}{3} \approx \frac{11 - 5.57}{3} \approx \frac{5.43}{3} \approx 1.81.
\]

### Step 6: Calculate Volume at \( x = 1.81 \)

Now substitute \( x = 1.81 \) back into the volume formula:
\[
V(1.81) = 1.81(10 - 2 \cdot 1.81)(12 - 2 \cdot 1.81).
\]

Calculating the dimensions:
- \( 10 - 2 \cdot 1.81 = 10 - 3.62 = 6.38 \)
- \( 12 - 2 \cdot 1.81 = 12 - 3.62 = 8.38 \)

So the volume is:
\[
V(1.81) = 1.81 \cdot 6.38 \cdot 8.38.
\]

Now calculating this:
1. Calculate:
- \( 6.38 \cdot 8.38 \approx 53.441 \)

2. Then:
- \( V(1.81) = 1.81 \cdot 53.441 \approx 96.749 \).

### Conclusion

Thus, when rounded to one decimal place, the maximum volume is approximately:

\[
\boxed{96.8 \text{ cubic inches}}.
\]

Thank you for your patience, and I apologize for the earlier miscalculation!