T = 5, so after 5 years
p(t) = t^3 - 14t^2 + 20t + 120
Take derivative to find minimum:
p’(t) = 3t^2 - 28t + 10
Factor to solve for t:
p’(t) = (3t - 2)(t - 5)
0 = (3t - 2)(t - 5)
0 = 3t - 2
2 = 3t
2/3 = t
Plug 2/3 into original equation, this is a maximum. We want the minimum:
0 = t - 5
5 = t
Plug back into original:
5^3 - 14(5)^2 + 20(5) + 120
125 - 14(25) + 100 + 120
125 - 350 + 220
- 225 + 220
p(5) = -5
Step-by-step explanation:
The outer angle at the top C of the ABC is 112 °. If the bisector of the side AB intersects the side AC at point Q and the segment BQ is perpendicular to AC, find the magnitude of ABC
Sin(63)=x/7
0.89100652418= x/7 > <span>0.89100652418 (7) = x/7(7)
</span><span>6.23704566932=x This is the height
</span>the area of a triangle is A=BH/2 or Base * Height /2
A= (<span>6.23704566932)(8)/2
</span>A=<span>24.9481826773 or 24.95 cm^2</span>
Answer:
70
Step-by-step explanation:
Angle Q and angle P are supplementary angles, so they add up to 180 degrees.
6x + 4 + 10x = 180
16x + 4 = 180
16x = 176
x = 176/16
x = 11
Put x in angle Q as 11 and solve.
6(11) + 4
66 + 4
= 70
Angle Q measures 70 degrees.
Answer:
<h2><u><em>
x=3 or x=−4</em></u></h2>
Step-by-step explanation:
when a number x is added to it's square, the total is 12. find two possible values of x
x+x^2=12
x^2+x=12
x^2+x−12=12−12
x^2+x−12=0
(x−3)(x+4)=0
x−3=0 or x+4=0
x=3 or x=−4
x=3 or x=−4
