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2 years ago

There are 100 books in the libary. There are non fiction and fiction books write the fraction of the books that are fiction

Mathematics

1 answer:

goblinko [34]2 years ago

6 0

Non fiction = x and fiction =y so x+y=100

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Round 6709 to the nearest thousand

8090 [49]

7000 is 6709 rounded to the nearest thousand

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2 years ago

Read 2 more answers

This is pretty easy :0

maksim [4K]

Answer:

1, 1.5, -1 1/5

Step-by-step explanation:

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3 years ago

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'Ciumane ananen<br> 32.50t + 5 = 28.75t + 20

gtnhenbr [62]

Answer:

The answer is 0.25

Step-by-step explanation:

(1)we group like terms by subtracting 32.50 by 28.75

(2) we subtract 20 and 5 to give us a value of 15

(3)when we subtract 32.50 and 28.75 we get a value of 3.75

(4)so the equation will look like this 3.75t = 15

(5)we divide 3.75 we both the two equations

(6)With a correct calculation, Your answer should be t = 0.25

6 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

2 years ago

Kevin says these number lines show 7/8 is equivalent to 4/5. Is Kevin correct?

notsponge [240]

Yes, Kevin is correct

3 0

2 years ago