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3 years ago

8

After learning how to fly, wendy reduced her daily commute time by 75%, percent. Previously, her commute took m minutes.what is

a expression that could represent her commute time in minutes after she learned how to fly?

2 answers:

Basile [38]3 years ago

5 0

The answer is:
Commute = m-m(0.75)

Tanya [424]3 years ago

4 0

That the answer. will be 0.7 because you put it ib similarly

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What is the answer to the equation -7x-3x+2= -8x - 8

timama [110]

Answer:

x=5

Step-by-step explanation:

-7x-3x+2= -8x - 8

Add like terms

-7x-3x=-10x

-10x+2=-8x-8

+8x +8x

-2x+2=-8

-2 -2

-2x=-10

Divide by -2 to isolate the variable

x=5

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2 years ago

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Ghella [55]

Answer:

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Step-by-step explanation:

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4 years ago

What is the most precise classification of the quadrilateral?

Naddik [55]

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3 years ago

Find the sum of the constants a, h, and k such that

ra1l [238]

Answer:

3

Step-by-step explanation:

The value of "a" is the coefficient of x^2, so we know that is 2.

__

<u>Solve for h</u>

Now, we have ...

2x^2 -8x +7 = 2(x -h)^2 +k

Expanding the right side gives us ...

= 2(x^2 -2hx +h^2) +k

= 2x^2 -4hx +2h^2 +k

Comparing x-terms, we see ...

-4hx = -8x

h = (-8x)/(-4x) = 2

__

<u>Solve for k</u>

Now, we're left with ...

2h^2 +k = 7 = 2(2^2) +k = 8 +k

Subtracting 8 we find k to be ...

k = 7 -8 = -1

__

And the sum of constants a, h, and k is ...

a +h +k = 2 +2 -1 = 3

The sum of the constants is 3.

6 0

4 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

4 years ago