Question

Please help:)

Answer 1

wat do you need a help with

Answer 2

The value of\[\Large{\frac{2^{m+3}\times3^{2m-n}\times5^{m+n+3}\times6^{n+1}}{6^{m+1}\times10^{n+3}\times15^m}}\]is??

Answer 3

@lgbasallote @mathslover Plz help:)

Answer 4

Use:

\[\huge a^x \times a^b = a^{x+y}\]

\[\huge \frac{a^x}{a^b} = a^{x-y}\]

Answer 5

I used but failed:(

Answer 6

Apply it to 6 first..

Answer 7

6=2x3
10=2x5
15=3x5

Answer 8

work on the denominator first

Answer 9

Sorry, but can I plz have full solution:)
I m so confused with this:(

Answer 10

plz plz

Answer 11

\[6^{m+1}×10^{n+3}×15^m=(2^{m+1}\times3^{m+1})\times(2^{n+3}\times5^{n+3})\times(3^m\times5^m)\]

Answer 12

\[\large \frac{2^m \times 2^3 \times 3^{2m} \times 3^{-n} \times 5^m \times 5^n \times 5^3 \times 6^n \times 6^1}{6^m \times 6^1 \times 2^n \times 2^3 \times 5^n \times 5^3 \times 3^m \times 3^5}\]

Answer 13

then @UnkleRhaukus

Answer 14

\[\large \frac{2^{m-n} \times 3^{m-n} \times 5^n \times 6^{n-m}}{3^5}\]

Answer 15

\[(2^{m+1}\times3^{m+1})\times(2^{n+3}\times5^{n+3})\times(3^m\times5^m)\]
\[=(2^{m+1+n+3})\times(3^{m+1+m})\times(5^{n+3+m})\]

Answer 16

\[\large \frac{5^n}{243}\]

Answer 17

Is this the answer..??

Answer 18

\[\Huge{\color{blue}{\cal{Answer:-}}}\]\[\LARGE{\color{green}{1.}}\]

Answer 19

\[\large \frac{2^m \times 2^3 \times 3^{2m} \times 3^{-n} \times 5^m \times 5^n \times 5^3 \times 6^n \times 6^1}{6^m \times 6^1 \times 2^n \times 2^3 \times 5^n \times 5^3 \times 3^m \times 5^m}\]

Answer 20

\[{\frac{2^{m+3}\times3^{2m-n}\times5^{m+n+3}\times2^{n+1}\times3^{n+1}}{2^{m+1+n+3}\times3^{m+1+m}\times5^{n+3+m}}}\]

Answer 21

I have some work PLz give me the ful ans @UnkleRhaukus @waterineyes @ash2326 @mathslover :)

Answer 22

I m offline

Answer 23

\[\huge \color{green}{ 2^{m-n} \times 3^{m-n} \times 6^{n-m} = 1}\]

Answer 24

\[={\frac{2^{m+3+n+1}\times3^{2m-n+n+1}\times5^{m+n+3}}{2^{m+1+n+3}\times3^{m+1+m}\times5^{n+3+m}}}\]
\[={\frac{\cancel{2^{m+3+n+1}}\times\cancel{3^{2m-n+n+1}}\times\cancel{5^{m+n+3}}}{\cancel{2^{m+1+n+3}}\times\cancel{3^{m+1+m}}\times\cancel{5^{n+3+m}}}}\]\[=\]

Answer 25

Firstly just separate all the terms by using the formula that i gave above:

\[\large \frac{2^m \times 2^3 \times 3^{2m} \times 3^{-n} \times 5^m \times 5^n \times 5^3 \times 6^n \times 6^1}{6^m \times 6^1 \times 2^n \times 2^3 \times 5^n \times 5^3 \times 3^m \times 5^m}\]

Just cancel the terms and again use the above formulas :

\[\large 2^{m-n} \times 3^{m-n} \times 6^{n-m} = 1\]

Answer 26

Oh! @Callisto @UnkleRhaukus @waterineyes thanx now I understood!

Answer 27

Welcome..

Answer 28

Everybody deserves a medal but I can give only 1 sorry:)

Answer 29

Really appreciate the help:)

Answer 30

As long as I can help, I'm very happy already. Medal doesn't mean that much to me!!

Answer 31

wow @Callisto