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marissa [1.9K]
3 years ago
11

Which ordered pair is a solution of the system of equations? y = 3x + 1 y = 5x - 1

Mathematics
1 answer:
kodGreya [7K]3 years ago
5 0

Answer:

y=3x+1=4

y=5x-1=4

Step-by-step explanation:

DO THE MATH

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8. What is the solution of the equation 4x + 12 = 48 ?
romanna [79]

Answer:

A) x=9

Step-by-step explanation:

48-12=36.

36/4=9

More detailed explanation:

4x+12=48

    -12    -12

     4x = 36

     /4     /4

     x   =    9

4 0
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R=9ft find the volume of each sphere   
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V = 4πr^{3} /3 = 972π;
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Here’s a table question only a few more table questions to come I need it in equation form
Sati [7]

Answer:

y = x

Step-by-step explanation:

Since the x-values and y-values remain the same all throughout the chart, it's in a 1 to 1 relationship.

That just means the numbers will ALWAYS EQUAL each other.

Hence, y = x

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3 years ago
Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified l
Sloan [31]

Answer:

The integral of the volume is:

V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

The result is: V = 78.97731

Step-by-step explanation:

Given

Curve: x^2 + 4y^2 = 4

About line x = 2 --- Missing information

Required

Set up an integral for the volume

x^2 + 4y^2 = 4

Make x^2 the subject

x^2 = 4 - 4y^2

Square both sides

x = \sqrt{(4 - 4y^2)

Factor out 4

x = \sqrt{4(1 - y^2)

Split

x = \sqrt{4} * \sqrt{(1 - y^2)

x = \±2 * \sqrt{(1 - y^2)

x = \±2 \sqrt{(1 - y^2)

Split

x_1 = -2 \sqrt{(1 - y^2)}\ and\ x_2 = 2 \sqrt{(1 - y^2)}

Rotate about x = 2 implies that:

r = 2 - x

So:

r_1 = 2 - (-2 \sqrt{(1 - y^2)})

r_1 = 2 +2 \sqrt{(1 - y^2)}

r_2 = 2 - 2 \sqrt{(1 - y^2)}

Using washer method along the y-axis i.e. integral from 0 to 1.

We have:

V = 2\pi\int\limits^1_0 {(r_1^2 - r_2^2)} \, dy

Substitute values for r1 and r2

V = 2\pi\int\limits^1_0 {(( 2 +2 \sqrt{(1 - y^2)})^2 - ( 2 -2 \sqrt{(1 - y^2)})^2)} \, dy

Evaluate the squares

V = 2\pi\int\limits^1_0 {(4 +8 \sqrt{(1 - y^2)} + 4(1 - y^2)) - (4 -8 \sqrt{(1 - y^2)} + 4(1 - y^2))} \, dy

Remove brackets and collect like terms

V = 2\pi\int\limits^1_0 {4 - 4 + 8\sqrt{(1 - y^2)} +8 \sqrt{(1 - y^2)}+ 4(1 - y^2)  - 4(1 - y^2)} \, dy

V = 2\pi\int\limits^1_0 { 16\sqrt{(1 - y^2)} \, dy

Rewrite as:

V = 16* 2\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

Using the calculator:

\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy = \frac{\pi}{4}

So:

V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

V = 32\pi * \frac{\pi}{4}

V =\frac{32\pi^2}{4}

V =8\pi^2

Take:

\pi = 3.142

V = 8* 3.142^2

V = 78.97731 --- approximated

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Its 1/3 thats the answer
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