I need help with this math problem.

The volume V of a pyramid with a square base with side length s and height h is V=1/3s^2h. Solved for s, this formula gives s= StartRoot StartFraction 3 Upper V Over h EndFraction EndRoot . A pyramid has a volume of approximately 7,500,000 cubic meters. Find the length of the side of the pyramid if its height is approximately 141 meters.
Here's a picture of the problem since I'm not able to write the equation out like its shown.

I need help with this math problemThe volume V of a pyramid with a square base with side length s and height h is V13s2h Solved for s this formula gives s Start class=

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Answer:

The answer to your question is s = 399.5 m

Step-by-step explanation:

Formula

                   [tex]s = \sqrt{\frac{3V}{h}}[/tex]

Data

V = 7 500 000 m³

h = 141 m

Substitution

                  [tex]s = \sqrt{\frac{3(7500000)}{141}}[/tex]

                  [tex]s = \sqrt{\frac{22500000}{141}}[/tex]

                 [tex]s = \sqrt{159574.47}[/tex]

                 s = 399.46 meters ≈ 399.5 meters

Answer:s = 400 meters

Step-by-step explanation:

The volume V of a pyramid with a square base with side length s and height h is

V=1/3s^2h

Solved for s, this formula gives

s = √3V/h

Given that the pyramid has a volume, V of approximately 7500000 cubic meters and the height is approximately 141 meters, the length if the side of the pyramid, s would be

s = √3 × 7500000/141

s = √3 × 53191.4894

s = √159574.4682

s = 399.468

s = 400 meters approximated to the nearest meter