Answer:
1/5^9
Step-by-step explanation:
You keep the 5 the same
Then you subtract the exponents...
-6 - 3 = -9
Since it’s negative and you can’t have a negative exponent you have to put a 1 as the numerator. So...
1/5^9
1/5 is equal to 2/10 so the one with 3/10 has more
I don't know if that is factor-able. It looks like you could just subtract and get 64t. which would lead to the answer being zero since you divide by zero when isolating the variable t.
Answer:
5in
Step-by-step explanation:
You can find it by subtracting the side that has 15 in and the side that contains that 10 in. 15-10=5
Since (f/g)(x) = f(x)/g(x) for x to be in the domain of (f/g)(x) it must be in the domain of f and in the domain of g. You also need to insure that g(x) is not zero since f(x) is divided by g(x). Thus there are 3 conditions. x must be in the domain of f: f(x) = 3x -5 and all real numbers x are in the domain of x.
Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o f )(x).
( f o f )(x) = f ( f (x))
= f (2x + 3)
= 2( ) + 3 ... setting up to insert the input
= 2(2x + 3) + 3
= 4x + 6 + 3
= 4x + 9
Given f(x) = 2x + 3 and g(x) = –x2 + 5, find (g o g)(x).
(g o g)(x) = g(g(x))
= –( )2 + 5 ... setting up to insert the input
= –(–x2 + 5)2 + 5
= –(x4 – 10x2 + 25) + 5
= –x4 + 10x2 – 25 + 5
= –x4 + 10x2 – 20
