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alekssr [168]
3 years ago
8

NEED HELP PLZZZZxxxx

Mathematics
1 answer:
nexus9112 [7]3 years ago
8 0
If you use rise/run, the answer is 3/4. Up 3, over 4
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What divided by 6 equals 7
vivado [14]

Answer:

42

Step-by-step explanation:

You can prove this by taking 42 and dividing it by 6, and you will see that the answer is 7. Tip: For future reference, when you are presented with a problem like "What divided by 6 equals 7?", all you have to do is multiply the two known numbers together.

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3 years ago
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Each Saturday, Tina mows lawns to earn extra money which she puts into a savings account. The graph shows the balance of Tina's
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2 years ago
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In-s [12.5K]
41 feet is the correct answer !!!!!!
6 0
3 years ago
Ray UW is the angle bisector of VUT.
Readme [11.4K]
<span>In the question "Ray UW is the angle bisector of VUT. If mVUW = (4x + 6)° and mWUT = (6x – 10)°, what is the measure of WUT?" It was given that UW is the angle bisector of VUT, which means that mVUW = mWUT. i.e. 4x + 6 = 6x - 10 6x - 4x = 6 + 10 2x = 16 x = 8 Therefore, the measure of WUT = 6x - 10 = 6(8) - 10 = 48 - 10 = 38 </span>
8 0
3 years ago
Read 2 more answers
Calculus 2 Master needed, evaluate the indefinite integral of: <img src="https://tex.z-dn.net/?f=%5Cint%5C%28%20%28lnx%29%5E2%7D
viva [34]

Answer:

\int (\ln(x))^2dx=x(\ln(x)^2-2\ln(x)+2)+C

Step-by-step explanation:

So we have the indefinite integral:

\int (\ln(x))^2dx

This is the same thing as:

=\int 1\cdot (\ln(x))^2dx

So, let's do integration by parts.

Let u be (ln(x))². And let dv be (1)dx. Therefore:

u=(\ln(x))^2\\\text{Find du. Use the chain rule.}\\\frac{du}{dx}=2(\ln(x))\cdot\frac{1}{x}

Simplify:

du=\frac{2\ln(x)}{x}dx

And:

dv=(1)dx\\v=x

Therefore:

\int (\ln(x))^2dx=x\ln(x)^2-\int(x)(\frac{2\ln(x)}{x})dx

The x cancel:

=x\ln(x)^2-\int2\ln(x)dx

Move the 2 to the front:

=x\ln(x)^2-2\int\ln(x)dx

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)

Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

=x\ln(x)^2-2\int 1\cdot\ln(x)dx

Let u be ln(x) and let dv be (1)dx. Thus:

u=\ln(x)\\du=\frac{1}{x}dx

And:

dv=(1)dx\\v=x

So:

=x\ln(x)^2-2(x\ln(x)-\int (x)\frac{1}{x}dx)

Simplify the integral:

=x\ln(x)^2-2(x\ln(x)-\int (1)dx)

Evaluate:

=x\ln(x)^2-2(x\ln(x)-x)

Now, we just have to simplify :)

Distribute the -2:

=x\ln(x)^2-2x\ln(x)+2x

And if preferred, we can factor out a x:

=x(\ln(x)^2-2\ln(x)+2)

And, of course, don't forget about the constant of integration!

=x(\ln(x)^2-2\ln(x)+2)+C

And we are done :)

8 0
3 years ago
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