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2 years ago

Given Angle 1 =xº and angle 2 = (x +50)°. Find the value of x. Then find angle 1 and 2

Mathematics

1 answer:

Elis [28]2 years ago

6 0

Answer:

x=65

Angle 1=65

Angle 2=115

Step-by-step explanation:

We know that angle 1 + angle 2 = 180. Writing it in terms of x, we get x + (x + 50)= 180. Then you have 2x+50=180. Subtracting 50 from both sides gives you 2x=130. Dividing both sides by 2, you get x=65. Angle 1 is 65 degrees. 180-65=115.

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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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2 years ago

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Yuliya22 [10]

Answer:

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2 years ago

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stira [4]

Answer:

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Step-by-step explanation:

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