Part A: f(t) = t² + 6t - 20
u = t² + 6t - 20
+ 20 + 20
u + 20 = t² + 6t
u + 20 + 9 = t² + 6t + 9
u + 29 = t² + 3t + 3t + 9
u + 29 = t(t) + t(3) + 3(t) + 3(3)
u + 29 = t(t + 3) + 3(t + 3)
u + 29 = (t + 3)(t + 3)
u + 29 = (t + 3)²
- 29 - 29
u = (t + 3)² - 29
Part B: The vertex is (-3, -29). The graph shows that it is a minimum because it shows that there is a positive sign before the x²-term, making the parabola open up and has a minimum vertex of (-3, -29).
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Part A: g(t) = 48.8t + 28 h(t) = -16t² + 90t + 50
| t | g(t) | | t | h(t) |
|-4|-167.2| | -4 | -566 |
|-3|-118.4| | -3 | -364 |
|-2| -69.6 | | -2 | -194 |
|-1| -20.8 | | -1 | -56 |
|0 | -28 | | 0 | 50 |
|1 | 76.8 | | 1 | 124 |
|2 | 125.6| | 2 | 166 |
|3 | 174.4| | 3 | 176 |
|4 | 223.2| | 4 | 154 |
The two seconds that the solution of g(t) and h(t) is located is between -1 and 4 seconds because it shows that they have two solutions, making it between -1 and 4 seconds.
Part B: The solution from Part A means that you have to find two solutions in order to know where the solutions of the two functions are located at.
Answer:
30t - 4
Step-by-step explanation:
Remember to follow PEMDAS. First, distribute -6 to all terms within the parenthesis:
2 -6(-5t + 1) = 2 + (-6 * -5t) + (-6 * 1)
2 + (30t) + (-6) = 2 + 30t - 6
Simplify. Combine like terms:
2 + 30t - 6 = 30t + (-6 + 2) = 30t - 4
30t - 4 is your answer.
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Tan 2 x-2 tan x=0
Tan x (tan x-2 ) =0
A tan x=0 —> x=0 and x =K π
B tan x=2 —> x=63.43 +k180 deg
I believe the lateral surface area is 87.96.
