how do you solve these two college algebra problems?:)?
a)
A company that manufactures bicycle frames has determined that the monthly profit (in dollars) from the sale of x frames is given by
P(x) = – 0.05x3 + 1.5x2 + 5x – 13000
Find the number of frames that must be sold per month for the profit to be zero. For what x-values will the profit be positive?
b)
The population of a certain endangered species of owl is declining exponentially. There are currently 450 living specimens, whereas just 8 years ago there were 12,000 . If the population continues to decline exponentially, how long will it be until there is only a single owl left?
i have answers but i want to know how to solve it:))
Comments
A)
Alright.
Assuming by x3, x2, you mean x^3 and x^2.
P(x) = -0.05x^3 + 1.5x^2 + 5x - 13000
Now, we want to know what causes it to be 0. Create an equation.
0 = -0.05x^3 + 1.5x^2 + 5x - 13000
Now, we need to solve for X, which looks...
...hard.
Alright, let's do this!
...or not.
Do your solutions look like this?
x = 400*(sqrt(3)*%i/2-1/2)/(3*(500*sqrt(1783067)/3^(3/2)-128500)^(1/3))+(500*sqrt(1783067)/3^(3/2)-128500)^(1/3)*(-sqrt(3)*%i/2-1/2)+10
x2 = (500*sqrt(1783067)/3^(3/2)-128500)^(1/3)*(sqrt(3)*%i/2-1/2)+400*(-sqrt(3)*%i/2-1/2)/(3*(500*sqrt(1783067)/3^(3/2)-128500)^(1/3))+10
x3 = (500*sqrt(1783067)/3^(3/2)-128500)^(1/3)+400/(3*(500*sqrt(1783067)/3^(3/2)-128500)^(1/3))+10
If so, leave whatever school you are attending, because it sucks.
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Assuming by x3, x2, you actually mean x3, x2
0 = -0.05x3 + 1.5x2 + 5x - 13000
Combine like terms
0 = 2.95x + 3.5x + 5x - 13000
0 = 11.45x - 13000
11.45x = 13000
x = 1135.37
The number of frames that must be sold per month is 1135.37.
For all X values greater than 1135.37, the profit will be positive.
Uhg. Now for B.
450 living specimens, 8 years, 12,000.
X = 12000
Y = 450
Z = 1
C = 8
First, we need to find the average rate of change.
(X-Y)/C should give us that. It is 1444.
Dividing 450 by 1444, we find that it will take around 0.311 years, or around 4 months and 3 days, to decline to 1.
for b)
12000=450+8*
I THINK.