Ask question

Ask question

All categories

3 years ago

15

Simplify the expression 6ab + 3ab - 7ab. What is the coefficient of the simplified expression?

2 answers:

Bad White [126]3 years ago

8 0

Simplifying 6ab + 3ab - 7ab
so we can write it
6ab + 3ab - 7ab = 0 ( because there is nothing on right side )
9ab - 7ab = 0
2ab = 0
So the coefficient of ab is 2.

Correct option is C.

loris [4]3 years ago

7 0

To simply to expression 6ab+3ab-7ab, we do:
6ab+3ab-7ab+= (Just do regular addition and subtraction if they have the same coefficent)
9ab-7ab=
2ab
So the answer is C.

You might be interested in

Which expression is equivalent to (82)−4

jok3333 [9.3K]

Answer:

82*-4 or

41*-8

Step-by-step explanation:

because math

4 0

3 years ago

Please help me with precal

Doss [256]

Consider the function $f(x)=\sqrt{x}.$ This function has:

the domain $x\in [0,\infty);$
the range $y\in [0,\infty).$

If the domain of unknown function is $[a,\infty),$ then $x\ge a$ or $x-a\ge 0.$ This means that you have $\sqrt{x-a}$ as a part of needed function.

If the range of unknown function is $[b,\infty),$ then $y\ge b.$ This means that you have to translate function b units up and then the expression of the function is

$y=\sqrt{x-a}+b.$

Answer: correct choice is B.

6 0

3 years ago

Goodmorning , please help!

Sav [38]

Answer:

7/2

Step-by-step explanation:

17=x+x+10

17-10=2x

7=2x

x=7/2

4 0

3 years ago

Read 2 more answers

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide

kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

<em>σ</em> = $60

<em>MOE</em> = $2

The critical value of <em>z</em> for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

$=[\frac{1.96\times 60}{2}]^{2}$

$=3457.44\\\approx 3458$

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0

3 years ago

It is exponent laws,I don't get it so can you go step by step plz

ipn [44]

I hope this helps you

5 0

3 years ago