Question

Find the product and simplify
4y/(y^2-81) × (y+9)/2y^2

Answer 1

2y3
y−9

Answer 2

Thats not one of the options ):

Answer 3

.....

Answer 4

@Learner11

Answer 5

2y3/y-9?

Answer 6

Thats not one of the choices..

Answer 7

sorry :(

Answer 8

Thats ok.

Answer 9

wait @ashley9337 . I think I can do it

Answer 10

Okay thank you !!

Answer 11

@imqwerty

Answer 12

Can you tag me when you figure it out? Thank you

Answer 13

oh! ok . sure :)

Answer 14

@imqwerty

Answer 15

@SyedMohammed98

Answer 16

\[\frac{ 2 }{ y(y-9) }\]

Answer 17

Thank you so much!

Answer 18

np :)

Answer 19

@imqwerty can you explain it to me ?

Answer 20

ok :)

Answer 21

@imqwerty can you help me with 3 more problems?

Answer 22

\[\frac{ 4y }{ y^2-81 }\times \frac{ y+9 }{ 2y^2 }\]
use the identity a^2 - b^2=(a+b)(a-b) in here - y^2 -81 u et-\[\frac{ 4y }{ (y+9)(y-9) }\times \frac{ y+9 }{2y^2}\]cut out the common things from numerator and denominator u get-\[\frac{ 2 }{ y(y-9) }\]

Answer 23

ok

Answer 24

What is the least commonn denominator for 3/x and 2/4xy

Answer 25

oh ! OK . Got it @imqwerty

Answer 26

least common divisor is the least common multiple.
its 4xy

Answer 27

Oh okay i understand. There a harder one that i have no idea about, find the lcd for x^2/(x^2+4x+3) and 2x/ (x^2+x-6)

Answer 28

@imqwerty

Answer 29

x^2+4x+3 --->(x+3)(x+1)

x^2+x-6------>(x+3)(x-2)

so the lcd wuld be (x+3)(x-2)(x+1)

Answer 30

Thank you for explaining it! okay can you help me with one last question

Answer 31

ok :)

Answer 32

Find the quotient and show all work please so i can understand how to do it :)
x^2-6x/7 ÷ x-6/x

Answer 33

ok :)

Answer 34

\[\frac{ x^2-6x }{ 7}\div \frac{ (x-6) }{ x}\]
we know that \[a \div b =a \times \frac{ 1 }{ b }\]
so we get -
\[\frac{ x^2-6x }{ 7 } \times \frac{ x }{ (x-6) }\]
take x common from x^2-6x u get->x(x-6)
\[\frac{ x(x-6) }{ 7 } \times \frac{ x }{ x-6 }\]\[\frac{ x^2 }{ 7 }\]

Answer 35

So the answer is x^2/7 right? Thank you so much for explaining it all.to me, i understand it now :)