All categories

2 years ago

Find f(h(-1)) 2 -9 11 -11

Mathematics

1 answer:

kumpel [21]2 years ago

5 0

Answer:

(d) -11

Step-by-step explanation:

Each of the functions is evaluated in the usual way: put the number where the variable is in the expression and simplify.

h(-1) = -2(-1)² -3(-1) +1 = -2 +3 +1 = 2

f(2) = (2(2) +7)/(2 -3) = 11/-1 = -11

Then f(h(-1)) is ...

f(h(-1)) = f(2) = -11

You might be interested in

Two less than two thirds times a number is ten

beks73 [17]

Let the number be x
2/3*x - 2 = 10
or, 2/3 * x =12
or x= 12*3/2
or, x = 18

6 0

3 years ago

Maddie does not have any pretzels. Zach gives Abbey 88 pretzels. Then he splits half of the pretzels he has left with Maddie. Za

Ann [662]

1212 x 2+ 88 = ?
Hope this helps

3 0

3 years ago

HELPP FOR BRIANLEST PLEASS

Alex777 [14]

Answer:

first one

Step-by-step explanation:

uh yeah the first one because it the second one to the original problem

6 0

2 years ago

Read 2 more answers

"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

The windows on a building make an array that has 4 rows. Each row has 3 windows

lutik1710 [3]

Answer:

12 windows

When there are arrays, you always do multiplication.

4 windows x 3 rows = 12 windows.

Hope it helps!

7 0

3 years ago