Y + 3 = 4(x - 5)
y + 3 = 4x - 20
y = 4x - 20 - 3
y = 4x - 23

There are two changes of sign, so there are 2 or 0 possible positive roots.

There are two changes of sign, so there are 2 or 0 possible negative roots.
There are 4,2 or 0 possible non-real roots.
Answer:
b. ![\sqrt[3]{4x+2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4x%2B2%7D)
Step-by-step explanation:
f(x) = ( x^3-2) /4
y = ( x^3-2) /4
Exchange x and y
x = ( y^3-2) /4
solve for y
Multiply each side by 4
4x = 4( y^3-2) /4
4x = y^3 -2
Add 2 to each side
4x+2 = y^3
Take the cubed root of each side
( 4x+2) ^ (1/3) = y^3 ^(1/3)
( 4x+2) ^ (1/3) =y
Answer:
When the price of a ticket = $0 or $40, there will be no revenue.
The revenue will be $500 if each ticket cost $37.3 or $2.68
Step-by-step explanation:
Let R(p) = p(200-5p)
when R(p) = 0
p(200-5p) = 0
p = 0 or 200-5p = 0
When the price of a ticket = $0 or $40, there will be no revenue.
when R(p) = 500
200p - 5p² = 500
5p² - 200p + 500 = 0
p = $37.3 or $2.68 (3 sig. fig.)
The required prices are $37.3 and $2.68