Question

Please answer this question I need this urgent Help meeeee

Answer 1

Answer:

$\frac{s}{2t}$

Step-by-step explanation:

When you have exponents above a like term and they are being multiplied together, you add them.

For example:

$a^{x} *a^{y} = a^{x + y}$

So let's group like terms in the numerator:

$4r^{3} r^{-5} s^{-2} s^{-1}t$   We can add terms like in the example.

$4 r^{-2} s^{-3} t$

Let's rearrange the denominator.

$8r^{-2} s^{-4} t^{2}$

Now we have:

$\frac{4r^{-2} s^{-3}t}{8 r^{-2} s^{-4} t^{2} }$  Cancel like terms

4/8 = 1/2    

$r^{-2} /r^{-2}$ = 1  So it cancels

$s^{-3} / s^{-4} = s^{-1}$ = s Since s is raised to the -1 it goes on top and becomes s.

$t / t^{2} = 1/t$

Now we combine everything back together:

$\frac{s}{2t}$