Answer:
x*$10 + y*$16 = $480
Step-by-step explanation:
The question seems incomplete, and after a search online I've found a lot of similar questions with similar numbers, so I really can not be sure of which one is the complete question.
I suppose that we want to get a linear equation that describes the given situation:
Let's define two variables:
x = number of hours that Maggie works in the office (for $10 each)
y = number of hours that Maggie works driving a tractor (for $16 each).
Then the total amount of money that Maggie gets is:
x*$10 + y*$16
Now we know that she earns $480 per week, then we know that:
x*$10 + y*$16 = $480
This is a linear equation we wanted to get, where (just to be complete) we need to add the restrictions:
x ≥ 0
y ≥ 0
(this means that she can not work a negative number of hours).
Howdy lol it’s 11:44pm over here where I’m at.
Answer:
A) $16
B) p(x) = 16x -800
C) 69 tickets
Step-by-step explanation:
A) The total of expenses is ...
$280 +100 +20 +400 = $800
If this is covered by 50 tickets, then a ticket must provide revenue of ...
$800/50 = $16
The cost per ticket is $16.
__
B) The profit is the difference between revenue and expenses. The revenue from sale of x tickets will be 16x. The expenses are fixed at 800, so the profit is ...
p(x) = 16x -800
__
C) We can find the number of tickets to sell (x) in order for profit to be at least $300 by solving the inequality ...
p(x) ≥ 300
16x -800 ≥ 300 . . . . . use the expression for p(x)
16x ≥ 1100 . . . . . . . . . add 800
x ≥ 68.75 . . . . . . . . . . divide by 16 . . . (the least satisfactory integer is 69)
In order to raise at least $300, the number of tickets sold must be at least 69.
The equation of this parabola will have the form f(x) = a(x+5)(x-4), which works out to f(x) = a(x^2 + x - 20). Since the parabola passes thru (3,40),
40 = a(3^2 + 3 - 20), or 40 = a(-8), so a = -5.
Thus, the equation of this parabola is y = -5(x^2 + x - 20).
