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2 years ago

Solve the equation-3y = 21 using inverse operations. What is the inverse operation step? What is the solution?

Mathematics

2 answers:

Ksju [112]2 years ago

6 0

Answer:

Y=-7

Step-by-step explanation:

GaryK [48]2 years ago

5 0

Answer:

-7

Step-by-step explanation:

You divide -3 on both sides to get the variable by itself, so -3 divided by 21 is -7

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Help w number 8, will mark brainliest !!!

eimsori [14]

Answer:

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because both the measurement and the congruent side are opposite in sentence

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2 years ago

The hanger diagram models the equation 2(n + 7) = 24. What could be the

baherus [9]

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first of all you times every number with two 2(n-7) then collect like terms then divide

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2 years ago

Which is more, 6 ounces or<br> 1 / 2<br> of a pound?

Llana [10]

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Answer: 1/2 a pound is more, because 1/2 a pound is 8 ounces

I hope this helped!

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5 0

2 years ago

Read 2 more answers

Simplify 7(n - p) + 5p.

andre [41]

7(n - p) + 5p
7n - 7p + 5p
7n - 2p

8 0

3 years ago

Read 2 more answers

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