Answer:
y = mx + b is the answer
Step-by-step explanation:
im a good teacher and i done this before
Answer:
f(x) = 4x^2 + 2x - 4.
Step-by-step explanation:
Let the quadratic function be y = f(x) = ax^2 + bx + c.
For the point (-2, 8) ( x = -2 when y = 8) we have:
a(-2)^2 + (-2)b + c = 8
4a - 2b + c = 8 For (0, -4) we have:
0 + 0 + c = -4 so c = -4. For (4, 68) we have:
16a + 4b + c = 68
So we have 2 systems of equations in a and b ( plugging in c = -4):
4a - 2b - 4 = 8
16a + 4b - 4 = 68
4a - 2b = 12
16a + 4b = 72 Multiplying 4a - 2b = 12 by 2 we get:
8a - 4b = 24
Adding the last 2 equations:
24a = 96
a = 4
Now plugging a = 4 and c = -4 in the first equation:
4(4) - 2b - 4 = 8
-2b = 8 - 16 + 4 = -4
b = 2.
<h3>
Answer: b = 4 and c = 7.</h3>
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Explanation:
Comparing y = x^2+bx+c to y = ax^2+bx+c, we see that a = 1.
The vertex given is (-2,3). In general, the vertex is (h,k). So h = -2 and k = 3.
Plug those three values into the vertex form below
y = a(x-h)^2 + k
y = 1(x-(-2))^2 + 3
y = (x+2)^2 + 3
Then expand everything out and simplify
y = x^2+4x+4 + 3
y = x^2+4x+7
We see that b = 4 and c = 7.
B and c, anything with an x^2 term is a quadratic.
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.