Find the prime factorization for the integer
121
this is what I think as far as I go
11x11
am I right?
3 answers
Right, but we usually write it as 11².
Simplify the expression. Assume all variables represent
nonzero real numbers.
-10b^5c^9/2b^5c^3
nonzero real numbers.
-10b^5c^9/2b^5c^3
Since it is a fraction, we look for like factors in the numerator and denominator and simplify accordingly.
Start with the coefficients:
(-10)/2 = -5
b5/b5 = 1
c9/c3 = c9-3 = c6
So putting it altogether:
-10b^5c^9/2b^5c^3
= -5c6
Start with the coefficients:
(-10)/2 = -5
b5/b5 = 1
c9/c3 = c9-3 = c6
So putting it altogether:
-10b^5c^9/2b^5c^3
= -5c6