Question

Is this right? please help me!! thank you <333

Simplify:

40v^2 over 35v^4 + 20v^3 over 5v

my answer: 7v^4 over 2

Answer 1

\[\frac{\frac{40v^2}{35v^4 + 20v^3}}{ 5v}\]\

Answer 2

or is it an addition?

Answer 3

could it be that the problem is actually

\[\Large \frac{40v^2}{35v^4} + \frac{20v^3}{5v}\]

Answer 4

yeah that is what i was wondering too

Answer 5

you are correct, @jim_thompson5910

Answer 6

\[\Large \frac{40v^2}{35v^4} + \frac{20v^3}{5v}\]

\[\Large \frac{40v^2}{35v^4} + \frac{7v^3*20v^3}{7v^3*5v}\]

\[\Large \frac{40v^2}{35v^4} + \frac{140v^6}{35v^4}\]


see where to go from here?

Answer 7

add both numirators?

Answer 8

correct

then simplify as much as possible

Answer 9

180

Answer 10

7/32?

Answer 11

no you can't add 40v^2 to 140v^6 since they aren't like terms

so you just leave it as

\[\Large \frac{40v^2+140v^6}{35v^4}\]

but you can still simplify this

Answer 12

32/7

Answer 13

this is because

\[\Large \frac{40v^2+140v^6}{35v^4}\]

becomes

\[\Large \frac{20v^2(2+7v^4)}{35v^4}\]

and that simplifies even further

Answer 14

my final answer is 2 over 7v^4

Answer 15

\[\Large \frac{40v^2+140v^6}{35v^4}\]


\[\Large \frac{20v^2(2+7v^4)}{35v^4}\]


\[\Large \frac{4(2+7v^4)}{7v^2}\]


\[\Large \frac{8+28v^4}{7v^2}\]

Answer 16

thats not an option....

Answer 17

what are your other options?