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2 years ago

Simplify 3^2 * 3^(-5). Express your answer as a fraction

Mathematics

1 answer:

murzikaleks [220]2 years ago

3 0

Answer:

1 /27

1 over 27

Step-by-step explanation:

brainliest please?

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1. Which picture below that does not contain enough information to prove that AABC = ADEF?

Viefleur [7K]

Answer:

The picture that does not contain enough information to prove that ΔABC = ΔDEF is

(3) Picture (3)

Step-by-step explanation:

The given information in picture (3) is the Angle-Side-Side of ΔABC corresponds with the Angle-Side-Side of ΔDEF,

However, the condition of Angle-Side-Side of ΔABC, is not sufficient to prove that ΔABC is congruent to ΔDEF congruency because the length of the unknown side can have two possible values

7 0

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:

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

5 0

2 years ago

Read 2 more answers

Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$