6 x 13/3=
78/3
divide top and bottom by 3
26/1
26
The square root of 18 is 4.2426, rounded down to the closest whole number is 4. Testing the integer values 1 through 4 for division into 18 with a 0 remainder we get these factor pairs: (1 and 18), (2 and 9), (3 and 6). The factors of 18 are 1, 2, 3, 6, 9, 18.
Step-by-step explanation:
Constant rate means its the same rate and didn't change over time
Answer:
250 batches of muffins and 0 waffles.
Step-by-step explanation:
-1
If we denote the number of batches of muffins as "a" and the number of batches of waffles as "b," we are then supposed to maximize the profit function
P = 2a + 1.5b
subject to the following constraints: a>=0, b>=0, a + (3/4)b <= 250, and 3a + 6b <= 1200. The third constraint can be rewritten as 4a + 3b <= 1000.
Use the simplex method on these coefficients, and you should find the maximum profit to be $500 when a = 250 and b = 0. So, make 250 batches of muffins, no waffles.
You use up all the dough, have 450 minutes left, and have $500 profit, the maximum amount.
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.
