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strojnjashka [21]
3 years ago
12

If you answer the question correctly I will give you brainliest

Mathematics
2 answers:
maxonik [38]3 years ago
8 0

Answer:

x = 21

Step-by-step explanation:

Cross multiply

4x = 8 * 10.5

4x = 84

x = 21

wel3 years ago
4 0

Answer:

x=21

Step-by-step explanation:

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-12×4÷6-4= how much is it​
Nat2105 [25]

Answer:

-12

Step-by-step explanation:

Hope this helps :D

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What are some ways in which companies can attract and retain employees? (site 2)
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The company can offer a variety of benefits not inly for the employee but also for the family. 
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3 years ago
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Over the last 3 months, Janet exercised a total of 101 hours. She exercised 4 hours more during the second month than the first
olganol [36]

In 1st month let she exercised h hours

in second month she exercised h+4 hours

in third month she exercised h+4-3=h+1 hours

Now

  • h+h+4+h+1=101
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2 years ago
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​A cable company has a one time installation fee and then a monthly charge to use their service. The total cost of modeled by th
SpyIntel [72]

Answer:

130=monthly charge

119=one time installation fee

Step-by-step explanation:

119 is the y-intercept and in problems like these, it is always the one time fee. 130 is the monthly payment because it has an x, which could account for the number of months. For example , say six months passed, you could multiply 130 * 6 to get the monthly fee for six months. I hope that makes sense!

6 0
2 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
3 years ago
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