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3 years ago

6

Which table represents a direct variation? Table A x 4 6 8 10 y 7 9 11 13 Table B x 4 6 8 10 y 12 18 24 30 Table C x 4 6 8 10 y

1 3 5 7 Table D x 4 6 8 10 y 3 3 3 3 Table A Table B Table C Table D

2 answers:

g100num [7]3 years ago

8 0

Answer: B: Table B

Step-by-step explanation: On Edgenuity!!

Alja [10]3 years ago

4 0

Answer:

Table B

Step-by-step explanation: skidaddle skidoodle you probably know the rest 8)

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X 1 2 3 4<br> y 5 10 15 20<br><br><br> What's the constant?

DiKsa [7]

<u>Answer:</u>

x ---> 1

y ---> 5

<u>Step-by-step explanation:</u>

We are given paired values for two variables x and y and we are to determine the constant number by which each term is increased such that they are in a proportional relationship.

For x, we have the following paired values:

$x: 1, 2, 3, 4$

So here the difference between each consecutive term is 1 so the constant is 1.

And for y, we have:

$y: 5, 10, 15, 20$

In this case, the difference between each consecutive term is 5 so the constant is 5.

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3 years ago

7th grade math help me please :))

Sav [38]

Answer:

no because the absolute value of -7 is 7 which is greater than the absolute value of 6 which is 6.

8 0

3 years ago

It took Sharon 4 hours to drive to her grandmother's house. Her grandmother lives 140 miles from Sharon.

Vladimir [108]

Answer:

You're right its B or 35

Step-by-step explanation:

Just divide 140 by 4 u get 35

3 0

3 years ago

Read 2 more answers

NEED HELP FAST<br> 30nPOINTS FOR BRAINLIEST<br><br><br> √5x + √6 = √9

vodka [1.7K]

$\sqrt{5x}$ + √6 = √9

The first step is to get $\sqrt{5x}$ by itself. We can do this by subtracting $\sqrt{6}$ from each side, and simplifying $\sqrt{9}$

$\sqrt{5x}$ = 3 - $\sqrt{6}$

Now we square both sides

5x = (3 - $\sqrt{6}$ )²

Using the formula (a - b)² = a² -2ab + b², (3 - $\sqrt{6}$ )² = 15 - 6 $\sqrt{6}$

5x = 15 - 6 $\sqrt{6}$

x = $\frac{15 - 6\sqrt{6}}{5}$

8 0

3 years ago

Read 2 more answers

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