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3 years ago

9

Y=X^2+5X+7 match each leading coefficient with its correct letter

1 answer:

Step2247 [10]3 years ago

3 0

Answer:

sdcfvgh

Step-by-step explanation:

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8 Line in the xy-plane contains points from each of Quadrants II, III, and IV, but no points from Quadrant I. Which of the follo

iren2701 [21]

Answer:

The correct option is D.

Step-by-step explanation:

The slope of a line is the change in y with respect to x.

$m=\frac{y_2-y_1}{x_2-x_1}$

If the slope of a line is undefined it means it is a vertical line and a vertical line can not passes through three quadrants. So, option A is incorrect.

If the slope of a line is 0 it means it is a horizontal line and a horizontal line can not passes through three quadrants. So, option B is incorrect.

If the slope of a line is positive it means the value of y increases as x increases.

Since it is an increasing line, therefore after a certain period both x and y will positive. It means the line will passes through first quadrant. So, option C is incorrect.

If the slope of a line is negative it means the value of y decreases as x increases. It can passes through each of Quadrants II, III, and IV.

Therefore the correct option is D.

6 0

3 years ago

I will give extra points please help

pav-90 [236]

Answer:

2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Can somone please help me 25.2÷7.8=

Gennadij [26K]

25.2 / 7.8 = ~ 3.23 hope that helps

5 0

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

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

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

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

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

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

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

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

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

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

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

Are the slopes 1/4 and -1/4 parallel, perpendicular, or neither?

anyanavicka [17]

-1/4 is not = -1 / 1/4 ( its = -4) so its neither .

6 0

3 years ago