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3 years ago

5

Please help!! Will give brainliest to the best answer!!

1 answer:

Andreas93 [3]3 years ago

8 0

The answer is H, you cannot make such conclusion

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A sample of students is taken from the school’s A honor roll. The school estimates that there are actually 360 students on the A

timurjin [86]

Answer:120.

Step-by-step explanation:this is because there are 3 grades at the school 6th 7th and 8th so you simple need to divide 360 by 3 and that gets you 360.

7 0

3 years ago

If ∆QRS ≈ ∆XYZ, then RS is congruent to?<br><br> A. XY<br> B. YZ<br> C. XZ<br> D. YX

Blababa [14]

Answer:

yz

Step-by-step explanation:

........

........

answer is YZ

7 0

3 years ago

Show that x²-8x+20 can be written in the form (x-a)²+a where a is an integer

Fiesta28 [93]

Answer:

x² - 8x + 20

= x²- 2*4*x + 16 + 4

= x²- 2*4*x + 4² + 4

= (x - 4)² + 4

= (x - a)² + a, with a = 4

Hope this helps!

5 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

I dont know how to find x

Verdich [7]

Simplify both sides:

x
—- = -9
4

now isolate the variable (x):

x= -9 • 4
x= -36

4 0

3 years ago