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3 years ago

PLEASE BE 100% SURE the answer choices are attached

Mathematics

1 answer:

DerKrebs [107]3 years ago

4 0

B) should be your answer

The first one (the one given), is 6 "points" (forgot what it's called sorry) from 0 on all sides. To dilate it 1/3, divide 6 and 3

6/3 = 2

The one's that have points that land on points with 2 in it is B

so B is your best answer

hope this helps

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The sum of person x and person y is 2640. if person x's income is 20% more than person y, find person x's income.

Masja [62]

Answer:

I do not understand she question do you mean the sum of their income is 2640

if so, then the answer is 440 to be x income

Step-by-step explanation:

x income = 20% or 1/5 of y income

y income = 5 of x income

we know that the sum of their income = 2640

which means that

x + y = 2640 =

1/5y + y = 2640

6y/5 = 2640

cross multiply to give us

6y = 5(2640)

6y = 13200

divide through by 6 to give us,

6y/6 = 13200/6

y = 2200

since we know that x income is 20% y income,

therefore, x income =

20/100 * 2200

= 440

therefore,

x income = 440

y income = 2200

5 0

2 years ago

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Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

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2 years ago

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maks197457 [2]

20:8, 130:52, 40:16, 190:76

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3 years ago

If f(x)=x+7 and g(x)=<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bx%20-%2013%7D%20" id="TexFormula1" title=" \frac{

Sergio [31]

Hi,

$f(x)=x+7\\

g(x)= \dfrac{1}{x-13} \\

(fog)(x)=f(g(x))=f(\dfrac{1}{x-13})=\dfrac{1}{x-13}+7\\


$
Domain of (fog)(x) is R \ {13} =(-oo 13[ U ]13 +oo)

8 0

2 years ago

I need help ASAP do anybody know how to do this??

Leviafan [203]

Answer:

f(0)=9

Step-by-step explanation:

All you do is plug in x=0 into the function.

In this case:

f(x)=10^x+8

f(0)=10^0+8

f(0)=1+8

f(0)=9

Therefore, f(0)=9

3 0

3 years ago

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