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yanalaym [24]
3 years ago
11

How do I find the percent and what is the answer DO NOT COMMENT THE ANSWER IF YOU CAN’T EXPLAIN

Mathematics
1 answer:
Marianna [84]3 years ago
4 0
64/100*75=48  That is the answer
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A family of three children and two adults visited a health club. The club charges $y an hour for each child and $x an hour for e
storchak [24]

Answer:

The first graph would be correct

Step-by-step explanation:

7 0
3 years ago
Hence solve the equation x^3+x^2-6x=0<br>​
Aneli [31]

<em>Answer:</em>

<em />

<em>x³ + x² - 6x = 0</em>

<em>x(x² + x - 6) = 0</em>

<em>x(x² + 3x - 2x - 6) = 0</em>

<em>x[x(x + 3) - 2(x + 3)] = 0</em>

<em>x(x - 2)(x + 3) = 0</em>

<em>x₁ = 0</em>

<em>x - 2 = 0 => x₂ = 2</em>

<em>x + 3 = 0 => x₃ = - 3</em>

4 0
3 years ago
Read 2 more answers
BRIANLIESTTT ASAP! PLEASE HELP ME :)
vagabundo [1.1K]

Answer:

A. \displaystyle observational\:study

Step-by-step explanation:

This takes alot of extended research before you even think about using the drug on patients.

I am joyous to assist you anytime.

7 0
3 years ago
What is 3 divided by what equals 36
Vedmedyk [2.9K]
3/x=36

Multiply both sides by x
3=36x

Then divide both sides by 36
3/36=x
1/12=x

Final answer: 1/12
6 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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