Question

Approximate √__8to the nearest hundredth. Show your work. Lesson 1-3
4. Solve the equation m2 = 14. Lesson 1-5
5. A fish tank is in the shape of a cube. Its volume is 125 ft3. What is the area
of one face of the tank? Lessons 1-4 and 1-5
6. Write 1.¯12 as a mixed number. Show your work.

Answers

Answered by GPT-4o mini
Sure, let's solve each of these problems step by step.

### 1. Approximate √8 to the nearest hundredth.

To approximate \(\sqrt{8}\), we can break it down as follows:

1. We know that \( \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2} \).
2. We can approximate \(\sqrt{2}\). It's approximately \(1.414\).
3. Now, multiply this by 2:
\[
2 \times 1.414 = 2.828
\]
4. Rounding to the nearest hundredth gives us:
\[
\sqrt{8} \approx 2.83
\]

### 2. Solve the equation \(m^2 = 14\).

To solve for \(m\):

1. Take the square root of both sides:
\[
m = \sqrt{14} \quad \text{or} \quad m = -\sqrt{14}
\]
2. The approximate value of \(\sqrt{14}\) can be calculated:
- Since \(3^2 = 9\) and \(4^2 = 16\), \(\sqrt{14}\) is between 3 and 4.
- A more precise approximation gives us \(\sqrt{14} \approx 3.742\).
3. Thus, the solutions are:
\[
m \approx 3.742 \quad \text{or} \quad m \approx -3.742
\]

### 3. Find the area of one face of a cube with volume \(125 \text{ ft}^3\).

To find the area of one face of the cube:

1. The volume \(V\) of a cube is given by \(V = s^3\), where \(s\) is the length of one side.
2. Setting the volume equal to \(125\):
\[
s^3 = 125
\]
3. Taking the cube root:
\[
s = \sqrt[3]{125} = 5 \text{ ft}
\]
4. The area \(A\) of one face of the cube is given by \(A = s^2\):
\[
A = 5^2 = 25 \text{ ft}^2
\]

### 4. Write \(1.\bar{12}\) as a mixed number.

To convert \(1.\bar{12}\) (where "12" repeats indefinitely) into a mixed number:

1. Let \(x = 1.\bar{12}\) or \(x = 1.121212...\).
2. To eliminate the repeating decimal, multiply by 100 (since "12" has 2 digits):
\[
100x = 112.121212...
\]
3. Subtract the original \(x\) from this equation:
\[
100x - x = 112.121212... - 1.121212...
\]
\[
99x = 111
\]
4. Solve for \(x\):
\[
x = \frac{111}{99}
\]
5. Simplify \(\frac{111}{99}\):
\[
\text{Both numerator and denominator can be divided by 3:} \quad \frac{111 \div 3}{99 \div 3} = \frac{37}{33}
\]
6. Now convert it to a mixed number:
- \(37 \div 33 = 1\) with a remainder of \(4\) (because \(37 - 33 = 4\)).
- So \(1 \frac{4}{33}\).

Therefore, the final answer for \(1.\bar{12}\) as a mixed number is:
\[
1 \frac{4}{33}
\]

In summary:
1. \(\sqrt{8} \approx 2.83\)
2. Solutions to \(m^2 = 14\) are \(m \approx 3.742\) or \(m \approx -3.742\)
3. Area of one face of the tank is \(25 \text{ ft}^2\)
4. \(1.\bar{12} = 1 \frac{4}{33}\) as a mixed number.
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