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3 years ago

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How do you solve this systems of equations problem 2r +3s=-5 6s=2r

1 answer:

Yuri [45]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Given the 2 equations

2r + 3s = - 5 → (1)

6s = 2r → (2) ( divide both sides by 2 ), then

r = 3s

Substitute r = 3s into (1)

2(3s) + 3s = - 5

6s + 3s = - 5

9s = - 5 ( divide both sides by 9 )

s = - $\frac{5}{9}$

Substitute this value into either of the 2 equations and evaluate for r

Substituting into r = 3s, then

r = 3 × - $\frac{5}{9}$ = - $\frac{5}{3}$

Solution is r = - $\frac{5}{3}$ and s = - $\frac{5}{9}$

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Volume:

$V \approx 888.02in^3 \\ \\ And, \ 750in^3$

<h2>Explanation:</h2>

A composite figure is formed by two or more basic figures or shapes. In this problem, we have a composite figure formed by a cylinder and a hemisphere as shown in the figure below, so the volume of this shape as a whole is the sum of the volume of the cylinder and the hemisphere:

$V_{total}=V_{cylinder}+V_{hemisphere} \\ \\ \\ V_{total}=V \\ \\ V_{cylinder}=V_{c} \\ \\ V_{hemisphere}=V_{h}$

So:

$V_{c}=\pi r^2h \\ \\ r:radius \\ \\ h:height$

From the figure the radius of the hemisphere is the same radius of the cylinder and equals:

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And the height of the cylinder is:

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The volume of a hemisphere is half the volume of a sphere, hence:

$V_{h}=\frac{1}{2}\left(\frac{4}{3} \pi r^3\right) \\ \\ V_{h}=\frac{1}{2}\left(\frac{4}{3} \pi (4)^3\right) \\ \\ V_{h}=\frac{128}{3}\pi in^3$

Finally, the volume of the composite figure is:

$V=240\pi+\frac{128}{3}\pi \\ \\ V=\frac{848}{3}\pi in^3 \\ \\ \\ V \approx 888.02in^3 \\ \\ And, \ 750in^3$

<h2>Learn more:</h2>

Volume of cone: brainly.com/question/4383003

#LearnWithBrainly

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