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Nimfa-mama [501]

3 years ago

14

Solve the inequality below. Use the drop-down menus to describe the solution and its greah

1 answer:

Solnce55 [7]3 years ago

6 0

Answer:

The solution of the inequality is x ≤ -4. A graph of the solution should have a vertical line passing through x = -4 and be shaded to the left of x = -4

Step-by-step explanation:

-7x + 13 ≥ 41

Subtract 13 from both sides

-7x ≥ 28

Dividing both sides by the -7 changes the inequality sign and we have

x ≤ -4

Hence, the solution of the inequality is x ≤ -4 and the graph of the solution should have a vertical line passing through x = -4 and it should be shaded to the left of x = -4 indicating that only numbers less than or equal to -4 are possible solutions of the inequality.

Hope this Helps!!!

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kolbaska11 [484]

Answer:

3 cm base, 6 cm height

Step-by-step explanation:

base is already labeled and the height is as well

3 0

4 years ago

Read 2 more answers

A gift box is a rectangular prism. The box measure 8 inches by 10 inches by 3 inches. What is it’s surface area?

MakcuM [25]

Answer:

the answer is 80 i think

Step-by-step explanation:

bc dependng on 2lw+2lh+2hw, formula 8 and 10 would be length and width

Lets say

8=width

10=lenght

3in=height

u would do

2 x l x w + 2 x l h + 2 x h x w=

2 x 10 x 8 + 2 x 10 x 3 + 3 x 8=

160 + 60 + 48+=

268 is the sa

pls brainliest and more points :0

8 0

4 years ago

(rs) (4) = (r/s) (3) =

Dmitriy789 [7]

Answer:

(rs) (4) = (r/s)(3) = (r)(s)(2)

4 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

HELP ill give brainliest to the RIGHT answer

labwork [276]

True im ded ok?........................

4 0

3 years ago