Answer:
This is hard to read, but i guess that the equation is:
[tex]7*5*2/7*3^2*5^0/2^{-3}*2^{-9}[/tex]
So we have some things to solve:
First, for any number:
x^0 = 1
then 5^0 = 1.
And for negative powers,
x^-n = (1/x)^n.
And we
Then we can rewrite our equation as:
[tex]\frac{7*5*2*3^2*1*2^3}{7*2^9}[/tex]
now we can cancel the seven in the numerator and the denominator
[tex]\frac{5*2*3*2^3}{2^9}[/tex]
now, remember the relation:
a^x*a^y = a^(x+y)
then:
2*2^3 = 2^4
and 2^4/2^9 = 2^-5 = (1^2)^5
Then our equation is:
[tex]\frac{5*3}{2^5}[/tex]
15/2^5 = 0.46875
