Question

@mathmath333

Answer 1

\(k^{-3}\) can be written as

\(\dfrac{1}{k^3}\)

Answer 2

and \(\dfrac{1}{k^{-3}}=k^3\)

Answer 3

first of all solve this term \(\rightarrow (ky^5)^2\)

Answer 4

can u simplify that

Answer 5

Yes, \[k^2y\] ^10

Answer 6

yes good

Answer 7

so \(\dfrac{(ky^2)^5}{k^{-3}y}=\dfrac{k^2y^{10}}{k^{-3}y}\)

now transfer \(k\) to denimainator and cancel the \(y\) in deniminator .

Answer 8

\[k^2*y\] ^10 ?

Answer 9

\(\dfrac{k^2y^{10}}{k^{-3}y}=k^2\cdot k^3y^{10-1}\)

simplify further

Answer 10

the power in \(k\) is incorrect check again \(2+3=\)

Answer 11

k^5

Answer 12

yes