Question

can someone help me plz.. i feel stupid because im so confused

Answer 1

what do u need help on?

Answer 2

Whatcha need?

Answer 3

I was not good enough :(
go ahead

Answer 4

you were great! @Ahmad , seriously, thank you.

Answer 5

Sorry @gabgurl , we're continuing a prior conversation. How can we help?

Answer 6

ok Simplify: the quantity 40 times a all over b squared minus 5 times b plus 4, end quantity all over the quantity 5 times a squared all over the quantity 2 times b minus 2, end quantity, end quantity. sorry

Answer 7

you are welcome @ybarrap, lol it seems this liitle missunderstand is very funny @gabgurl @ybarrap

Answer 8

lol (:

Answer 9

\[\frac{ \frac{ 40a }{ (b-1)(b-4) } }{ \frac{ 5a*a }{ 2(b-1) } }\]

Answer 10

okay so far @gabgurl

Answer 11

\[\huge \frac{\frac{40a}{(b^2-5)(b+4)}}{\frac{5a^2}{2(b-2)}}\] this?

Answer 12

yes except the -5 and b are together. and the 2 and the b but the rest is right

Answer 13

yup but the 2 and b r together

Answer 14

\[\huge \frac{\frac{40a}{(b^2-5b+4)}}{\frac{5a^2}{2b-2)}}\]

Answer 15

hmm may be I went too fast :P

Answer 16

yuppp thats right

Answer 17

$$\tt \frac{\dfrac{40a}{b^2-5b+4}}{\dfrac{5a^2}{(2b-2)}}$$

Answer 18

yes

Answer 19

\[\large \frac{40a}{b^2-5b+4}\cdot \frac{2b-2}{5a^2}\]\[\large \frac{8\cancel{40a}}{(b-4)\cancel{(b-1)}} \cdot \frac{2\cancel{(b-1)}}{a\cancel{5a^2}}\]

Answer 20

\[\large \frac{2(8)}{a(b-4)}\] simplify :)

Answer 21

thank yas (: now i understand......... thank u @Jhannybean

Answer 22

no problemo :) And thank you!

Answer 23

no problems (:

Answer 24

$$\tt \frac{\dfrac{40a}{b^2-5b+4}}{\dfrac{5a^2}{(2b-2)}}\\
=\frac{4a}{b^2-5b+4}\frac{2b-2}{5a^2}\\
=\frac{4a}{5a^2}\frac{2(b-1)}{(b-1)(b-4)}\\
=\frac{8}{5a}\frac{1}{b-4}$$

Answer 25

thank u!! (:

Answer 26

*all times 10

Answer 27

its 40a, not 4a.

Answer 28

you're right, that's why I said *all times 10.

Answer 29

oh i see :)

Answer 30

but still, multiplying by numerator and denominator by 10 wouldnt give you the right answer..... \[\large \frac{2(8)}{a(b-4)} = \frac{16}{a(b-4)} \]\[\large 10 \left(\frac{8}{5a(b-4)}\right) = \frac{80}{50a(b-4)} \]\[\large \frac{16}{a(b-4)}\ne \frac{80}{50a(b-4)}\]

Answer 31

yup thats wrong.. but the answer u got was right jhanny and thank u again(:

Answer 32

$$\tt \frac{\dfrac{40a}{b^2-5b+4}}{\dfrac{5a^2}{(2b-2)}}\\
=\frac{4a}{b^2-5b+4}\frac{2b-2}{5a^2}\\
=\frac{4a}{5a^2}\frac{2(b-1)}{(b-1)(b-4)}\\
=\frac{8}{5a}\frac{1}{b-4}\\
*Correction(should~ have ~started ~with ~40, ~not~ 4):\\
=\frac{10(8)}{5a}\frac{1}{b-4}\\
=\frac{2(8)}{a}\frac{1}{b-4}\\
$$

Answer 33

Good job :)

Answer 34

lol