a/b * b/c * c/d * d/e is equal to a/e provided that b, c, d,
and e are not zero
PROVE
a/b * b/c * c/d * d/e
= (a/b *b/c) * (c/d * d/e)
= ab/bc * (c/d * d/e)
= a/c * (c/d * d/e)
= a/c * (cd/de)
= a/c * c/e
= ac/ce
= a/e
Therefore, a/b * b/c * c/d * d/e is equal to a/e provided that
b, c, d, and e are not zero
Just lol, it up and the you will be fine.
The answer is 5.
You just plug whatever number is in the f(?) part in for x!
9514 1404 393
Answer:
- 90 student tickets
- 65 adult tickets
Step-by-step explanation:
For many "mixture" problems, it is convenient to use a variable for the quantity of the highest contributor. Here, we can use 'a' to represent the number of adult tickets, because adult tickets cost the most. Then the number of student tickets is (a+25), and the total revenue is ...
6a +3(a+25) = 660
9a +75 = 660
9a = 585
a = 65
(a+25) = 90
The drama class sold 90 student tickets and 65 adult tickets.
(g/f)(x)= g(x)/f(x) =
= (x² - 6) /(3x +1)
3x+1≠0
3x≠-1
x≠ - 1/3
(g/f)(x)= (x² - 6) /(3x +1), when x≠ - 1/3
Answer: A.
