Mathematics: 8. the base of a circular cone has a diameter of 10 cm and an altitude of 10 cm. the cone is filled with water. a sphere is lowered into the cone until it just fits. exactly one-half of the sphere remains out of the water. once the sphere is removed, how much water remains in the cone?

8. the base of a circular cone has a diameter of

Ответ:

mimacapa

08.01.2022

The volume of water that remains on the cone is 523.6 cm³

Step-by-step explanation:

To solve this problem you have to keep in mind the formules that describes the volume of a cone and the volume of a sphere.

Volume of a cone = (πr²h)/3

Volume of a sphere = (4/3)πr³

So, if the base of the cone has a diameter of 10 cm, its radius is 5 cm. Its altitude is 10 cm. ⇒Volume = (πr²h)/3 ⇒ Volume = [π(5²)10) ⇒

Volume = 785.4 cm³. This is the initial volume of water.

Now if the sphere fits in the cone and half of it remains out of the water, the other half is inside the cone. Estimating the volume of the sphere and dividing it by two, you find the volume of water that was displaced.

Volume of a sphere = (4/3)πr³, here the radius is the same of the base of the cone (5 cm).

⇒ Volume = (4/3)π(5³) ⇒ Volume = 523.6 cm³ ⇒ The half of this volume is 261.8 cm³. This is the volume of water displaced.

⇒ The volume of water that remains on the cone is 523.6 cm³ (785.4 cm³- 261.8 cm³)

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Ответ:

ZachLaVine2016

23.06.2021

Mathematics

I'm thinking you want something that looks like this
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$ or it could be like this
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$

use chain rule
dy/dx f(g(x))=f'(g(x))g'(x)
and √x=x^(1/2)
so
I'll do the first one first
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$ =
$\frac{dy}{dx}(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}$ =
$\frac{1}{2(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}}(1+sin(2x)cos(2x))$ =
$\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}$

the 2nd one is a bit simpler
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$ =
$\frac{dy}{dx}x^{\frac{1}{2}}+\frac{1}{4}sin^2(2x)$ =
$\frac{1}{2}x^{\frac{-1}{2}}+\frac{1}{4}sin^2(2x)$ =
$\frac{1}{2x^{\frac{1}{2}}}+sin(2x)cos(2x)$ =
$\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)$

in conclusion
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$ = $\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}$ and
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$ = $\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)$
depends which one it was

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8. the base of a circular cone has a diameter of 10 cm and an altitude of 10 cm. the cone is filled with water. a sphere is lowered into the cone until it just fits. exactly one-half of the sphere remains out of the water. once the sphere is removed, how much water remains in the cone?

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