Dose anyone know any websites for big ideas answers ?

What is a simplified fraction of -4/17?

Zarrin [17]

I think there is no simplified fractions for this number it will be the same -4/17

4 0

2 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

List P contains m numbers; list Q contains n numbers. If the two lists are combined to produce list R, containing m + n numbers,

Anarel [89]

Answer: YES

Step-by-step explanation:

We need to write out the expressions

P= {m}

Q= {n}

R= {m+n}

If 2m=n then we can say;

P= {½n} Q= {n} & R= {³/²n}

It is obvious that the smaller number in Q is greater than the largest number in P

We can make some assumptions.

Let n= (x,y,z)

Consequently,

P={½x,½y,½z} Q={x,y,z} and R= {1.5x,1.5y,1.5z}

Therefore the median will be the middle element,

Median of P= ½y

Median of Q = y

Median of R = 1.5y

And 1.5y>1.5y

Then we can agree that the median of R is greater than the median of both P and Q

5 0

3 years ago

WELL, SO FAR THIS SITE IS FAILING ME BAD UNFORTUALETY! DAMMIT! I NEED SMART PEOPLE WITH RIGHT ANSWERS AND NOT INCOMPLETE OR WRON

Mandarinka [93]

Answer:

How can I help you?

Step-by-step explanation:

I will try my hardest and best to help you.

(p.s. I just saw your profile and most of the questions include pictures, on my end, it says that these pics are blocked. So, try asking the question without having to take a picture, instead, just type the question out. (and include the choices, if there are any))

3 0

3 years ago

Please help me guys, its five questions

ruslelena [56]

Answer:

1. A 2.d 3.b 4.a 5. c

Step-by-step explanation:

7 0

3 years ago