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2 years ago

How to solve 21 = m +14 1/6

Mathematics

2 answers:

murzikaleks [220]2 years ago

7 0

Answer:

Inverse operations by subtracting 14 1/6 from each side of the equation

Step-by-step explanation:

You need to first isolate the variable (leave only the variable on 1 side of the equal sign and everything else on the other). To do this do inverse operations (to get rid of "14 1/6" you need to subtract 14 1/6) from both sides of the equation (from both sides so the equation remains equal).

21-14 5/6 = 6 5/6

This leaves us with the solution

m=6 5/6

Oduvanchick [21]2 years ago

5 0

Answer:

$\frac{56}{3}$

Step-by-step explanation:

Image

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Given that Segment BE is 5 units, find the indicated measures for each

zalisa [80]

Answer:

The indicated measure of the length of AE = 5 units

Step-by-step explanation:

To determine:

What is the indicated measure of length?

Information Fetching and Solution Steps:

From this single diagram, it seems the indicated measure of length which we have to determine is the length of AE.

Given that Segment BE is 5 units. i.e. BE = 5 units

As the square has two equal diagonals AC and DB which meet at point E.

So, point E being the common mid-point of both diagonals.

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And since both diagonals AC and DB are equal in length, it means the length of AC must be 10 units.

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Therefore, the indicated measure of the length of AE = 5 units

Keywords: square, diagonals, length, line segment

Learn more about square diagonals from brainly.com/question/628263

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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:

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

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3 years ago

ANSWER ASAP PLEASE ONLY IF CORRECT!

taurus [48]

Need more detail to solve

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