-4(3m−3)+ (−10+8m)=?

1 answer:

7 0

-4m+2
Is your equation you must now answer
An easy way to add and subtract unlike terms is to think of
Same sign
Add and keep
Different sign
Subtract and keep the sign of the bigger number
Also known as SADS

You might be interested in

A print shop orders 58 T-shirts. They order 16 pink T-shirts, and six times as many blue T-shirts as white T-shirts. How many bl

Pepsi [2]

The have ordered 36 blue T-shirts.
___________________

How I got it --- So, we take away 16 from 58. You get 42.
________________________________________

So, from then, you know 36 is 6 times as many as 6, right? ( Because 6 x 6 = 36)
___________________________________________________________

So, then, you have ordered 36 blue shirts and 6 white shirts, because 36 + 6 =42.
______________________________________________________

Glad I could help, and good luck!
_________________________
_________________________
_________________________

3 0

2 years ago

Read 2 more answers

A circle with a radius of 5 inches would have an area of about what?

Marrrta [24]

The area of a circle of radius r is pi*r^2

Hence here the area is A=5^2*pi=25pi~78.5 square inches

6 0

2 years ago

Read 2 more answers

Which of the following statements are true about the equation below?

dangina [55]

The correct answer is 5

8 0

2 years ago

Read 2 more answers

Lena says that and are like terms. Is she correct? Why or why not? A. No, because the same variables are not raised to the same

qwelly [4]

Answer:

D because of the question

8 0

2 years ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

2 years ago