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3 years ago

The sum of two consecutive interegers is -131. Find the larger number.

1 answer:

6 0

$a,a+1- two\ consecutive\ integers\\\\
a+a+1=-131\\\\
2a+1=-131\ \ \ | subtract\ 1\\\\2a=-132 \ \ \ | divide\ by\ 2\\\\
a=-66\\\\
Numbers\ are:\ -66,-65.$

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Solve the equation using multiplication or division.

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Answer:

y=-4

Step-by-step explanation:

-7y=28

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28/-7=-4

Answer=-4

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9 months ago

A bag contains 5 black marbles, 7 blue marbles, 3 green marbles, 4 red marbles, and 1 white marble. A marble is selected and the

schepotkina [342]

Answer:

2/25

Step-by-step explanation:

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2 years ago

If (x+2) is a Factor x^3 + 2x^2 + 2x + k then find the value of K.

Delvig [45]

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Step-by-step explanation:

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2 years ago

Read 2 more answers

2 candy bars cost $.98 at this rate how much do 5 candy bars cost? I NEED ASAP

alisha [4.7K]

Answer:

2.45$

Step-by-step explanation:

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7 0

2 years ago

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 :)

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2 years ago