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Anika [276]
3 years ago
13

A hollow ball is made of rubber that is 2 centimeters thick. The ball has a radius to the outside surface of 6 centimeters. What

is the approximate volume of rubber used to make the ball? Use 3.14 for pi. 33.5 cm³ 267.9 cm³ 636.4 cm³ 904.3 cm³
Mathematics
1 answer:
forsale [732]3 years ago
7 0
Your answer is: 636.4 cm³
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-3 \cdot f(-8) + 7 \cdot g(2) =−3⋅f(−8)+7⋅g(2)=minus, 3, dot, f, left parenthesis, minus, 8, right parenthesis, plus, 7, dot, g,
dsp73

Answer:

-22

Step-by-step explanation:

The graph from which the information is to be read is attached below.

We want to find the value of −3⋅f(−8)+7⋅g(2).

From the graph:

  • f(-8)=-2
  • g(2)=-4

Therefore:

−3⋅f(−8)+7⋅g(2)=−3(-2)+7(-4)

=6-28

=-22

−3⋅f(−8)+7⋅g(2)=-22

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2 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
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\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
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3 years ago
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