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vlada-n [284]
4 years ago
15

HELP!!!!!!!!!!!!!!!!!!!!!!ASAP!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Gwar [14]4 years ago
7 0
Best answer I can find is 10%
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What is the slope of the line that passed through (3,-4) and (4,9)
Advocard [28]

Step-by-step explanation:

<u>Step 1:  Find the slope</u>

<u />m=\frac{y_2-y_1}{x_2-x_1}

m=\frac{9-(-4)}{4-3}

m =\frac{9+4}{1}

m=13

Answer: The slope is 13

5 0
4 years ago
20 POINTS An electrician charges a set fee for every house call and then charges an hourly rate depending on how long the job ta
tatuchka [14]

Answer:

the \: slope \: is \to \boxed{60}

Step-by-step explanation:

if \to C=65+60t \\  C=60t + 65 \\ but \: the \: \boxed{ c - intercept} \: is \: in \: the \: form \to \\ c = mt + b \\ hence \to \\ 60t = mt \\ m =  \frac{60t}{t}  \\ m = 60 \\  the \: slope \: is \to \boxed{60}

5 0
3 years ago
Read 2 more answers
Consider the first month's sales data for the Midwest region.
myrzilka [38]

Answer:

1. Significantly different from

1. Inconsistent

Step-by-step explanation:

Midwest region total stores equal to 63.1%. The percentage for current store is less than total stores therefore data in the given table is not consistent. The data given is relatively for the first month sales and is significantly different from the other stores.

8 0
3 years ago
What does the quantity 28-8 represent?
nikklg [1K]
You question implies simple subtraction is needed:

28-8=20
5 0
2 years ago
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
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