In other words, the radius of cylinder 1 is the same as that of cylinder 2, and the height of cylinder 2 is the same as that of cylinder 1. This is what it means for two solid shapes to be congruent.

Now, to find the volume of cylinder 2, we apply the frmula for the volume of a tyical cylnder:

$\text{Volume}=\pi\times r^2\times h$

Now,

radius of cylinder 2 = radius of cylinder 1 = 3

height of cylinder 2 = 8

Thus: r = 3, h= 8

Therefore:

$\begin{gathered} \text{Volume}=\pi\times r^2\times h \\ \text{Volume}=\pi\times3^2\times8 \\ \text{Volume}=\pi\times9^{}\times8 \\ \text{Volume}=\pi\times72 \\ \text{Volume}=72\pi\text{ cubic units} \end{gathered}$

Therefore the volume of cylinder 2 is 72pi cubic units (Option D)

6 0

1 year ago

Consider the initial value problem 2ty′=4y, y(2)=−8. 2ty′=4y, y(2)=−8. find the value of the constant cc and the exponent rr so

Softa [21]

2ty'=4y
Replacing y'=dy/dt in the equation:
2t(dy/dt)=4y
Grouping terms:
dy/y=4dt/(2t)
dy/y=2dt/t
Integrating both sides:
ln(y)=2ln(t)+ln(c), where c is a constant
Using property logarithm: b ln(a) = ln(a^b), with b=2 and a=t
ln(y)=ln(t^2)+ln(c)
Using property of logarithm: ln(a)+ln(b) = ln(ab), with a=t^2 and b=c
ln(y)=ln(ct^2)
Then:
y=ct^2
Using the initial condition: y(2)=-8
t=2→y=-8→c(2)^2=-8→c(4)=-8
Solving for c:
c=-8/4
c=-2

Then the solution is y=-2t^2
Comparing with the solution: y=ct^r
c=-2, r=2

Answer: T<span>he value of the constant c is -2 (c=-2) and the exponent r is 2 (r=2)</span>

5 0

3 years ago

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