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White raven [17]
3 years ago
13

EASY question for y’all mathy people, easy points! Question in photo.

Mathematics
2 answers:
bija089 [108]3 years ago
4 0
I think the answer is D
OverLord2011 [107]3 years ago
3 0

Answer:

Step-by-step explanation:

A linear function is a line which, in slope-intercept form, follows

y = mx + b,

where m is the slope and b is the y-intercept.

The only one that follows that form is B. B can be rewritten as

y=\frac{1}{3}x+2

and in that form we see that the slope value is 1/3 and the y-intercept is 2. The choice is B.

A is a radical function moved up from its originating spot; C is an inverse function moved down 4 units from its originating spot with a vertical asymptote at x = 0; and D is a cubic moved down 10 units from its originating spot.

B is the one you want.

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3 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

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