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3 years ago

The product of 18 and Q algebra expression

Mathematics

2 answers:

zavuch27 [327]3 years ago

7 0

Answer:

$18 \times q$

$18q$

multiply 18 by q to get 18q as the answer

Nady [450]3 years ago

3 0

Answer:

18q

Step-by-step explanation:

18*q=18q

Hope that helped :)

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A student drove to the university from her home and noted that the odometer reading of her car increased by 16.0 km. The trip to

notka56 [123]

Answer:

Average speed is 48 km per hour

Step-by-step explanation:

A student drove to the university from her home and noted that the odometer reading of her car increased by 16.0 km. The trip took 20.0 min. (a) What was her average speed? in km/hr

Distance traveled is 16 km and the time taken is 20 minutes

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Help find x. I believe it’s 13

aksik [14]

Answer:

m∠RST = 155°

m∠RSU = 102°

Step-by-step explanation:

m∠RST = m∠RSU + m∠UST

12x - 1 = 9x - 15 + 53

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3 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Therefore,

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

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3 years ago