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3 years ago

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of the total number of pages in a book, 62% have pictures and the rest have no pictures. There are 310 pages with pictures in th

e book. What is the total number of pages in the book?

1 answer:

Dmitry [639]3 years ago

4 0

The answer is that: there is 62% of 310? The answer is 186 pages.

You might be interested in

What number do you divide by 12 to get 9

finlep [7]

Well I don't know. Let's figure it out together.

You said (a number) divided by 12 gives you 9 .

A fraction is the easiest way to show division, so
you can write this equation:

   (number) / 12 = 9

Now you can multiply each side of the equation by 12 .
When you do that, you have ...

   number = 9 x 12

   <em>number =    108</em> .

Now you know how to find it. That's even better than just
having the answer.

5 0

3 years ago

Read 2 more answers

Alex and Freddy collect stamps. The number of stamps in Alex’s collection is in ratio 3:5 to the number of stamps in Freddy’s co

Tems11 [23]

Suppose Alex has 3x stamps, Freddy has 5x stamps. After Freddy gives Alex 10 stamps, Alex has 3x+10 while Freddy has 5x-10
the new ratio is 7 to 9, so we have the equation: (3x+10)/(5x-10)=7/9
9(3x+10)=7(5x-10)
27x+90=35x-70
160=8x
x=20
So Alex has 3*20=60 stamps while Freddy has 5*20=100. Together they have 160

4 0

2 years ago

The ages of several trees in a neighborhood are shown below: 27 years, 3 years, 2 years, 25 years, 5 years What is the mean age

nata0808 [166]

Mean is equal to sum/ number of values

27+3+2+25+5=62
62/5=12.4

Final answer: D, 12.4 years

5 0

2 years ago

Read 2 more answers

HELP PLZ PICTURE^ Need it asap porfa

timurjin [86]

Answer:

it has one solution

Step-by-step explanation:

1.y=x-3

2. 3y-3x sub x-3 in place of y therefore

it can also be written as 3x-3x-9=9

if you add 9 to both sides 3x-3x-9+9=-9+9

0+0=0

3 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