Math  /  Geometry

QuestionDetermine the side length of each cube to the nearest tenth of a unit.
Square 1 \square Select an answer Square 2 \square Select an answer

Studdy Solution

STEP 1

What is this asking? We need to find the side length of two cubes, one with a volume of 6307 mm36307 \text{ mm}^3 and another with a volume of 24411 ft324411 \text{ ft}^3, rounded to the nearest tenth. Watch out! Don't forget to apply the cube root correctly and keep track of the units!

STEP 2

1. Cube 1 Side Length
2. Cube 2 Side Length

STEP 3

Alright, let's **start** with the first cube!
We know that the volume of a cube is given by V=s3V = s^3, where VV is the **volume** and ss is the **side length**.
We're given the volume, so we need to solve for ss.

STEP 4

To **isolate** ss, we'll take the cube root of both sides of the equation.
This gives us V3=s33\sqrt[3]{V} = \sqrt[3]{s^3}.
Since the cube root "undoes" the cubing operation, we get s=V3s = \sqrt[3]{V}.

STEP 5

Now, let's **plug in** the given volume of the first cube, which is V=6307 mm3V = 6307 \text{ mm}^3.
So, s=63073s = \sqrt[3]{6307}.

STEP 6

Using a calculator, we find that 6307318.472 mm\sqrt[3]{6307} \approx 18.472 \text{ mm}.

STEP 7

Finally, we need to **round** to the nearest tenth.
Since the hundredths digit is 77, which is greater than or equal to 55, we round up.
So, the side length of the first cube is approximately 18.5 mm18.5 \text{ mm}.

STEP 8

Now, let's **tackle** the second cube!
We'll use the same formula: s=V3s = \sqrt[3]{V}.

STEP 9

This time, the volume is V=24411 ft3V = 24411 \text{ ft}^3.
So, s=244113s = \sqrt[3]{24411}.

STEP 10

Using a calculator, we find that 24411329.012 ft\sqrt[3]{24411} \approx 29.012 \text{ ft}.

STEP 11

Rounding to the nearest tenth, since the hundredths digit is 11, which is less than 55, we round down.
So, the side length of the second cube is approximately 29.0 ft29.0 \text{ ft}.

STEP 12

The side length of the first cube is approximately 18.5 mm18.5 \text{ mm}.
The side length of the second cube is approximately 29.0 ft29.0 \text{ ft}.

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