Question

2πr^2 * 2πr + 2r

Answer 1

Just started pre calc... this is the first problem in the book and I have no idea :( ... Not looking for just an answer.. it looks so easy yet I have no idea

Answer 2

Isn't there some kind of question asked about it in the book? Or a certain calculation?

Answer 3

Exactly what do you need to do?

Answer 4

It would help if we had the chapter this was under.

Answer 5

well all it says is to simply each expression. Its for Functions Modeling Change - A preparation for Calculus. Chapter 1 problem 3. Its an algebra problem and this is simply a skill refresher so no explanation..

Answer 6

YOu could try to do the multiplication and then factor out the common factors.

Answer 7

Ok, so I here we go, multiplication leads me to 4πr^2 + 2πr^2 so 6πr^4

Answer 8

6πr^4 ?... does that sound anywhere near

Answer 9

No, because \[2πr^2 * 2πr + 2r=4\pi^2r^3+2r\]

Answer 10

omg
I messed uo

Answer 11

2πr^2 + 2πr * 2r

Answer 12

Thats the actual problem... Sorry

Answer 13

Now you've got two not-alike terms:\[4\pi^2r^3\]and\[2r\]

So you *cannot* add them!
You can, however, write the whole thing as a product, by factoring out the common factors:\[4\pi^2r^3=2r \cdot 2\pi^2r^2\]
\[2r= 2r \cdot 1\]So 2r is a common factor. You can write it as follows now:\[4\pi^2r^3+2r=2r \cdot2\pi^2r^2+2r \cdot 1=2r(2\pi^2r^2+1)\]

Answer 14

ok ok, I think I got it. So \[2\pi r ^{2} \times 2\pi r + 2r\]

we start with multiplication.

So I get \[4\pi ^{2} * r ^{3} + 2r\]

I would say that about as good as its going to get or it needs to be ? The rest of it sort of confuses me but thats considered simplified correct?

The correct problem... which I messed up on, is actually \[2\pi r ^{2} + 2\pi r * 2r\]

This one would be:

multiplication on the left side of the equation first so:
\[4\pi r^{2} \]
\[2\pi r^{2} +4\pi r^{2} \]
That would give me
\[6\pi r^{2} \]

Answer 15

Then I have another one... being
\[\frac{12 \pi - 2\pi }{ 6 \pi }\]

I would say,
\[\frac{2 \pi (6 - 1) }{ 6 \pi }\]

Then

\[\frac{5 }{ 3 \pi }\] OR
\[\frac{5 \pi }{ 3 \pi }\]

Not sure which one would be correct?

Answer 16

In the numerator, you have 10π, so after dividing by 6π, you have 10/6 = 5/3

Answer 17

wow... well that was a dumb one, I over complicate this stuff :/ ... thank you!

Answer 18

yw!

Answer 19

this is weird... c + 1/2c = 2/3c according to the back of the book... but
Isnt it c + 1/2c = 1c + 1/2c = 1 1/2 c

Answer 20

Are you sure it doesn't say 3/2 c?

Answer 21

Positive! ... my teacher complains about how this book is the worst book he's ever used... so that should explain that.