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3 years ago

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

3 years ago

Read 2 more answers

R=9ft find the volume of each sphere

Vedmedyk [2.9K]

V = 4π $r^{3} /3$ = 972π;

6 0

3 years ago

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

6 0

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

3 0

3 years ago

ANSWER PLS :D SAY TOP LEFT BOTTOM LEFT TOP RIGHT BOTTOM RIGHT OR JUST<br> SAY THE ANSWER

Damm [24]

Its 1/3 thats the answer

5 0

3 years ago