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.
1) 4
2) 16
3) 16
4) 6
5)12
6) 4
Answer:
We want to simplify:
(3 + 1/4)*(3/5)
The first step is to write the first term as a single rational number.
We know that:
3*1 = 3
and 4/4 = 1
then:
3*1 = 3*(4/4) = (3*4)/4 = 12/4
We do this because we want to have the same denominator in both numbers, so we can directly add them.
Then we get:
(3 + 1/4)*(3/5) = (12/4 + 1/4)*(3/5) = (13/4)*(3/5)
And remember that in the multiplication of rational numbers the numerator are multiplied together and the same for the denominators, then we get:
(13/4)*(3/5) = (13*3)/(4*5)
If we solve the multiplications we get:
(13*3)/(4*5) = (39/20)
Now, we can notice that in the numerator we have two prime numbers, 13 and 3.
And in the denominators, we have a 4 (which is equal to 2*2) and a 5.
So the prime numbers in the numerator and the denominator are all different, this means that we can not simplify it furthermore.
Then we have:
(3 + 1/4)*(3/5) = (39/20)
(-3,2)(1,2)...notice that the y values are the same.....when the y values are the same u have a horizontal line with a 0 slope.
IF the x values would have been the same (instead of the y values), then u would have had a vertical line with an undefined slope.