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3 years ago

A store employes 11 women and 12 men. What percentage of employes are men?

Mathematics

1 answer:

pishuonlain [190]3 years ago

8 0

Answer:

12%

Step-by-step explanation:

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Solve for x leave answer as decimal <br> 6/5=x/3

777dan777 [17]

Answer:

3 3/5 =x

3 .6 =x

Step-by-step explanation:

6/5 = x/3

Use cross products

6*3 = 5*x

18 = 5x

Divide each side by 5

18/5 = 5x/5

18/5 =x

3 3/5 =x

8 0

3 years ago

Read 2 more answers

Ms. Ball is applying for a credit card. The one she wants to get has a 7% annual interest rate, compounded semiannually, and cha

jeka94

Answer:

I need points

Step-by-step explanation:

4 0

2 years ago

What is equivalent to k/2

Mazyrski [523]

Answer:

k x 1/2

Step-by-step explanation:

because k x 1/2 = k/2

8 0

3 years ago

Dan multiplied 1,563 by 4 and got the incorrect product of 2,252. Find the correct product and explain what Dan did wrong. Pleas

Morgarella [4.7K]

Answer:

The correct product is 6,252.

I don't know what Dan did wrong lol

8 0

3 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago