If you factor the denominator of the left multiplicand...
[(a-4)(a+3)]/[a(a-4)] you can see that the (a-4)s cancel out leaving:
(a+3)/a
now you have
(a+3)/a * 2a^3/((a+3)(a-1)) so the (a+3)s cancel out leaving:
1/a * 2a^3/(a-1) which is:
2a^3/(a(a-1)) so now a^1 cancel leaving
2a^2/(a-1)
Set the composite results function.
Evaluate f(g(x)) by substituting the value of g into f.
Simplify each term.
Apply the distributive property.
Multiply 4 by 2.
Multiply 2 by 3.
Subtract 5 from 6.
There are at least 3 houses on a street if at least two of them have addresses that are consecutive integers.
No, there is not common ratio...
0.972 (common ratio is 0.3...10.8*.3*.3)
you can use the geometric mean to find this center value easily...
(18*8)^(1/2)=12
Solution
-3b + 11 = -18 + 8b
-3b - 8b = -18 - 11
-11b = -29
b = 29/11
