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Vadim26 [7]
3 years ago
5

If some one could help with this it would be awesome!!

Mathematics
1 answer:
natulia [17]3 years ago
3 0

Check the picture below.

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What is -3(x+9)=15?
Vadim26 [7]
Answer is x= -14 because you divide each term by -3 and then simplify
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Help please!!!! ASAP WILL MARK BRAINLIST IF REALLY FAST
Setler79 [48]

Answer:

1. Product

2. Difference

3. Difference

Step-by-step explanation:

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3 years ago
Two times a number added to another number is 25. Three times the first number minus the second number is 20. What are both numb
lawyer [7]

Answer:

the two number are 7 and 9

Step-by-step explanation:

2*x+y=25

3x-y = 20

add the two equations

2*x+y=25

3x-y = 20

---------------

5x = 45

divide by 5

5x/5 = 45/5

x = 9

lets find y

2x+y = 25

substitute x

2*9 +y =25

18+y =25

subtract 18 from both side

y = 25-18

y =7

7 0
4 years ago
Solve and graph the absolute value inequality: |2x + 1| ≤ 5. number line with closed dots on -3 and 2 with shading going in the
Andrei [34K]
We are given with the inequality |2x + 1| ≤ 5 and asked to solve the equation. In this case, we take first the positive side, that is 2x + 1 ≤ 5. this is equal to 2x ≤ 4 or x ≤ 2. For the negative side, the equality is -5 ≤ 2x + 1. This is equal to -6 ≤ 2x or -3 ≤ x. Hence the solution is  -3 ≤ x ≤ 2. The answer is B. closed dots on -3 and 2 with shading in between. The equal in <span>≤  means closed dots.</span>
6 0
3 years ago
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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}

5 0
3 years ago
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