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SIZIF [17.4K]
3 years ago
15

Willie earns $7.00 per hour working. He

Mathematics
2 answers:
My name is Ann [436]3 years ago
7 0

Answer:

$420

Step-by-step explanation:

Find how much he earned by multiplying 7 by the number of hours he worked

7(60)

= 420

So, Willie earned $420

Alex777 [14]3 years ago
3 0

Answer:

He earned $420.

Step-by-step explanation:

I hope this helps! Have a great rest of your day!

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What is 100 times less than one million​
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5. The number −6 is 2 more than an unknown number. Find the unknown number.
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-4

Step-by-step explanation:

-6 - 2 = -4

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What is the equation for the line of reflection? On a coordinate plane, triangle A B C has points (6, 3.7), (5.4, 2), (1, 3). Tr
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Answer:

The equation for the line of reflection will be x = y.

Step-by-step explanation:

On a coordinate plane, triangle Δ ABC has points (6, 3.7), (5.4, 2), (1, 3). Triangle Δ A'B'C' has points (3.7, 6), (2, 5.4), (3, 1).

And we have to find the equation of the line of reflection of the points.

Now, it is clear from the coordinates of vertices of Δ ABC and Δ A'B'C' that after reflection the x and y-values of the respective coordinates interchange and there is no change in signs.

Therefore, the equation for the line of reflection will be x = y. (Answer)

4 0
2 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}

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