We have that
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>
Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.
Answer:
g(h(x)) = [x + 3]^2 or x^2 + 6x + 9
Step-by-step explanation:
g[h(x)] signifies that h(x) is the input to g(x).
Writing out g(x) = x^2 and replacing "x" with [x + 3], we get:
g(h(x)) = [x + 3]^2 or x^2 + 6x + 9
Answer:
A
Step-by-step explanation:
+7 means go up
if the graph was 7x, the graph enlarges by a factor of 7
Did u type that by mistake
You have to combine like terms, so the variable (x, y, s, d, c....) and the exponents must be the same in order to combine them.
For example:
x² + x³ Since they don't have the same exponent, you can't combine them
y² + 3y² = 4y²
23x + x = 24x
4. 2s² + 1 + s² - 2s + 1 You can rearrange it if it makes it easier
2s² + s² - 2s + 1 + 1 = 3s² - 2s + 2
5. 5t² - 2t - 1 - (3t² - 5t + 7) Distribute/multiply the - to (3t² - 5t + 7)
5t² - 2t - 1 - 3t² + 5t - 7 = 2t² + 3t - 8
Do the same for #9 and #10, and you should get:
9. 2k² + 5k - 9
10. 6y³ - 7y² - 6y - 12