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.
You said that 'A is 30' right there in the equation
so it would be
Xsquared=30
Answer:
C
Step-by-step explanation:
C is the only one that is an enlarged version of A.
