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cupoosta [38]
3 years ago
8

What number multiplied by itself 3 times equals -64

Mathematics
1 answer:
charle [14.2K]3 years ago
7 0
-64/3 divide -64 by 3

-21.33333333333333333
or
-21 1/3
You might be interested in
PLEASE HELP WILL GIVE BRAINLIEST
zalisa [80]

If we look at the rest of the input and output, we also realize that:

         output = input + 5

Thus if the input is 'n' then the output is 'n+5'

Hope that helps!

8 0
2 years ago
|6x|+3=21 solve for X (absolute value)​
Fudgin [204]

Answer:

x=3\text{ or } x=-3

Step-by-step explanation:

So we have the equation:

|6x|+3=21

First, subtract 3 from both sides:

|6x|=18

Definition of absolute value:

6x=18\text{ or } 6x=-18

Divide both sides by 6 for both equations:

x=3\text{ or } x=-3

And we're done!

6 0
3 years ago
Read 2 more answers
If the population of a certain city increased 25% in two years, the new population was what percent of the old?
disa [49]

Answer:

125%

Step-by-step explanation:

This because if in origin the city was made up for example by 100,000 people, that was the 100% of the city. Now you add the 25% in two years and the city will be made by 125,000 people. So <em>100%+25%=125%</em><em>.</em><em> </em>

I think it's right

7 0
3 years ago
Plzzzz help me with this problem
hichkok12 [17]

Answer:

Option B

Step-by-step explanation:

The Europeans moved to American colonies in order to increase their wealth and broaden their influence over world affairs.

7 0
3 years ago
Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis
Tasya [4]
Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5


A.
\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx

B.
sepearte the integrals
\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}

next one
\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}

the last one you can do yourself, it is \frac{50}{3}
the sum is \frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}


so the area under the curve is \frac{233}{3}
6 0
3 years ago
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