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alexandr402 [8]
1 year ago
10

The sphere is _____ cubic centimeters bigger than the cube. (Round to the nearest cubic centimeter.)

Mathematics
1 answer:
Misha Larkins [42]1 year ago
6 0

ANSWER

The sphere is 10762 cubic centimeters bigger than the cube.

EXPLANATION

We want to find the difference in the volumes of the sphere and the cube.

To do this, we have to find the volumes of the sphere and cube and subtract that of the cube from the sphere.

The volume of a sphere is given as:

V=\frac{4}{3}\pi r^3

where r = radius

The radius of the sphere is 15 centimeters. Therefore, the volume of the sphere is:

\begin{gathered} V=\frac{4}{3}\cdot\pi\cdot15^3 \\ V\approx14137\operatorname{cm}^3 \end{gathered}

The volume of a cube is given as:

V=s^3

where s = length of the side

The length of the side of the cube is 15 centimeters. Therefore, the volume of the cube is:

\begin{gathered} V=15^3 \\ V=3375\operatorname{cm}^3 \end{gathered}

Therefore, the difference in the volumes of the sphere and cube is:

\begin{gathered} V_d=V_s-V_c \\ V_d=14137-3375 \\ V_d=10762\operatorname{cm}^3 \end{gathered}

Therefore, the sphere is 10762 cubic centimeters bigger than the cube.

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<em>               ⇒ choose</em> two sets of coordinates and find the slope

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Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ
atroni [7]

Answer: y=Ce^(^3^t^{^9}^)

Step-by-step explanation:

Beginning with the first differential equation:

\frac{dy}{dt} =27t^8y

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

\frac{1}{y} \frac{dy}{dt} =27t^8

Multiply both sides by 'dt' to get:

\frac{1}{y}dy =27t^8dt

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt

ln(y)=27(\frac{1}{9} t^9)+C

ln(y)=3t^9+C

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)

y=e^(^3^t^{^9} ^+^C^)

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

y=e^(^3^t^{^9}^)e^C

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

y=Ce^(^3^t^{^9}^)

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8

Now check if the derivative equals the right side of the original differential equation:

(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)

Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

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