How do you find the inverse of a function?

Mathematics

1 answer:

pashok25 [27]3 years ago

3 0

This or just google it

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In the parallelogram below find why

Mkey [24]

Angle E=130degrees because those two are congruent. Therefore if you 180-130 you'll get the other angle in the triangle with 70degrees and y. 180-130=50degrees. Then add 50 and 70. 50+70=120degrees. A triangle is supposed to have 180 degrees inside.
180-120=60degrees.
Therefore y=60degrees

8 0

3 years ago

a. If a rabbit can move 7/5 miles every hour, then how many hours would it take for a rabbit to go 7 miles?

sergeinik [125]

5 hours because 7/5 *5/1=35/5 and 35 divided by 5 equals 7

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4 years ago

Kyra is using rectangular tiles of two types for a floor design. A tile of each type is shown below: Two rectangular tiles, rect

Rudiy27

Since there is no picture shown, I just plotted the given data points myself. That is shown in the attached picture. The blue rectangle is rectangle PQRS while the orange one is rectangle JKLM. I believe there are some choices for this question but you forgot to include. Nevertheless, I will give my observations from the given figure.

The tile PQRS is bigger than tile JKLM. A rectangle is a two-dimensional shape that has two sets of equal parallel planes. Thus, its area is equal to the length multiplied by its width.

tile PQRS = (9-5)*(12-7) = 20 units²
tile JKLM = (6-4)*(10-5) = 10 units²

Tile PQRS is larger by 10 units².

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3 years ago

Form the greatest and the smallest 4-digit number using the following digits

MrMuchimi

Greatest =9998

Smallest=8889

Thanks!

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3 years ago

Integrate e^x(sin(x) cos(x))

Karo-lina-s [1.5K]

$I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx$

$\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}$

Looking for $I_2$ :

$\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I$

Replacing:

$2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}$

6 0

3 years ago