Question

Reduce this algebraic fraction.

Answer 1

yikes
guess we have to factor

Answer 2

yea i needed help with that. its so frustrating i almost started to cry lol

Answer 3

\[\frac{3(4a^2-1)\times (2a+1)}{2(2a+1)^2\times 6}\] is a start
want to see how i got that or is it clear?

Answer 4

yes please

Answer 5

ok lets work them step by step (btw that is not the answer, that is just step one)

Answer 6

i know :)

Answer 7

first off you can factor out a 3 from \(12a^2-3\) right? you get
\[3(4a^2-1)\]

Answer 8

yes

Answer 9

as for \((2a+1)^{-2}\) the \(-2\) exponent means it goes in the denominator with an exponent of \(2\)

Answer 10

so far we have
\[\frac{3(4a^2-1)}{(2a+1)^2}\]

Answer 11

alright makes sense so far

Answer 12

then \[\left(\frac{6}{2a+1}\right)^{-1}\] the \(-1\) in the exponent means flip it giving
\[\frac{6}{2a+1}\]

Answer 13

ok

Answer 14

and there was a 2 in the denominator originally
that is how i ended up with
\[\frac{3(4a^2-1)\times (2a+1)}{2(2a+1)^2\times 6}\]

Answer 15

oh damn made a huge mistake, sorry to confuse you

Answer 16

lol...yea i was likw huh

dont worry about its fine

do u still know how to do it?

Answer 17

what i meant to say was right, but i wrote it wrong
\[\left(\frac{6}{2a+1}\right)^{-1}\] the minus sign in the exponent means flip it, so it is
\[\frac{2a+1}{6}\] doh

Answer 18

that is how i ended up with
\[\frac{3(4a^2-1)\times (2a+1)}{2(2a+1)^2\times 6}\]

Answer 19

i dont understand where in the equation that goes

Answer 20

ohh wait nvm

Answer 21

it is all a multiplication
you multiply fractions straight across

Answer 22

\[\frac{3(4a^2-1)}{2}\times \frac{1}{2a+1}\times \frac{2a+1}{6}=\frac{3(4a^2-1)\times (2a+1)}{2(2a+1)^2\times 6}\]

Answer 23

now some little more factoring

Answer 24

\[4a^2-1=(2a+1)(2a-1)\] as the difference of two squares

Answer 25

I see

Answer 26

that makes \[\frac{3(4a^2-1)\times (2a+1)}{2(2a+1)^2\times 6}\] in to
\[\frac{3(2a+1)(2a-1)\times (2a+1)}{2(2a+1)^2\times 6}\]

Answer 27

now we cancel our brains out

Answer 28

you have two factors of \(2a+1\) in the numerator, and also two in the denominator

Answer 29

yes

Answer 30

we cancel those yea

Answer 31

\[\frac{3(\cancel{2a+1)}(2a-1)\times \cancel{(2a+1)}}{2\cancel{(2a+1)^2}\times 6}\]

Answer 32

leaving
\[\frac{3(2a-1)}{12}\] oh and you can cancel more
\[\frac{2a-1}{4}\]

Answer 33

Thank you, imt aking notes on everything you wrote lol

Answer 34

lots of cancellation here, rarely happens like this in real life
no crying in math, just ask here

Answer 35

yw

Answer 36

thank you very much!

Answer 37

:)

Answer 38

are you done or you got more?

Answer 39

i do actually have another

Answer 40

you want to do it with help or by yourself?

Answer 41

if it is geometry i am lost

Answer 42

its algebra II

Answer 43

displacement of cylinders?

Answer 44

Yes

Answer 45

hope they give you a formula to work with
post a screen shot might be easiest

Answer 46

Alright, good idea

Answer 47

lol we can do this
what a stupid question

Answer 48

Lol ok

Answer 49

lets go step by step

Answer 50

okie

Answer 51

oops it is \(bore =\frac{r}{2}\) not matter we get
\[\frac{bore}{2}=\frac{r}{4}\]

Answer 52

i see

Answer 53

that makes
\[\left(\frac{r}{4}\right)^2=\frac{r^2}{16}\]

Answer 54

yes

Answer 55

stroke = \(h\) and \(n=8\)

Answer 56

so
\[D=\pi\frac{r^2}{16}\times h\times 8\]

Answer 57

cancel the 8 leaves
\[C=\frac{\pi r^2 h}{2}\]

Answer 58

cept it is D not C

Answer 59

ohh i see.

Answer 60

thank you :)

Answer 61

that rly helped me

Answer 62

yw
you good with plugging in the next numbers?

Answer 63

yea i can take it from here thanks :)

Answer 64

yw