Question

help pleaze

Answer 1

\(\large \frac{4a^4b^2c}{12a^2b^{-5}c^3}\)

Answer 2

lets simplify the numbers first

Answer 3

4 goes in 12 right ?

Answer 4

yes 3times

Answer 5

good,

\(\large \frac{\cancel{4}a^4b^2c}{\cancel{12}^3a^2b^{-5}c^3}\)

Answer 6

ok whats next

Answer 7

\(\large \frac{a^4b^2c}{3a^2b^{-5}c^3}\)

Answer 8

next simplify \(a\)

Answer 9

a^2

Answer 10

Yes !

\(\large \frac{a^2b^2c}{3b^{-5}c^3}\)

Answer 11

next simplify the varibale \(b\)

Answer 12

\(\large \frac{a^2b^7c}{3c^3}\)


(why 7 ?)

Answer 13

next simplify the variable \(c\)

Answer 14

c^3

Answer 15

check again, top c is there

Answer 16

it wud becoem :-

\(\large \frac{a^2b^7}{3c^2}\)

Answer 17

but its exponent is 1 right

Answer 18

exactly thats why we reduce 1 in the bottom \(c^3\) becomes \(c^2\)

Answer 19

ohhh

Answer 20

below how it happens :-

\(\large \frac{a^2b^7c}{3c^3}\)

\(\large \frac{a^2b^7c^1}{3c^3}\)

\(\large \frac{a^2b^7}{3c^{3-1}}\)


\(\large \frac{a^2b^7}{3c^2}\)

Answer 21

oh i get it

Answer 22

see if that makes sense... if u dont get it fully, its okay... as you do more problems, you will see these im sure :)

Answer 23

ok