The maximum diameter of a bowling ball is 8.6 inches. What is the volume to the nearest tenth of a bowling ball with this diameter?

The volume of the bowling ball to the nearest tenth is v ≈ 106.0π in³ in terms of π  or v≈332.9 in³

Step-by-step explanation:

To find the volume of the bowling ball, we will follow the steps below;

First write down the formula for calculating the volume of a sphere

v=πr³

where v = volume of the ball   and  r = radius of the ball

From the equation given, the diameter of the ball is 8.6 inches but radius is half of diameter, this implies that r= d/2 = 8.6/2 = 4.3, that is; radius=4.3 inches.

Let substitutes the value into the equation

v=×π×(4.3)³

v = ×π×79.507

v =106.009π

v ≈ 106.0π in³

we can take π=3.14

v = 106.009×3.14

v =332.86826

v≈332.9 in³

The volume of the bowling ball to the nearest tenth is v ≈ 106.0π in³ in terms of π  or v≈332.9 in³

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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