The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2
<h3>How to determine the functions?</h3>
A quadratic function is represented as:
y = a(x - h)^2 + k
<u>Question #6</u>
The vertex of the graph is
(h, k) = (-1, 2)
So, we have:
y = a(x + 1)^2 + 2
The graph pass through the f(0) = -2
So, we have:
-2 = a(0 + 1)^2 + 2
Evaluate the like terms
a = -4
Substitute a = -4 in y = a(x + 1)^2 + 2
y = -4(x + 1)^2 + 2
<u>Question #7</u>
The vertex of the graph is
(h, k) = (2, 1)
So, we have:
y = a(x - 2)^2 + 1
The graph pass through (1, 3)
So, we have:
3 = a(1 - 2)^2 + 1
Evaluate the like terms
a = 2
Substitute a = 2 in y = a(x - 2)^2 + 1
y = 2(x - 2)^2 + 1
<u>Question #8</u>
The vertex of the graph is
(h, k) = (1, -2)
So, we have:
y = a(x - 1)^2 - 2
The graph pass through (0, -3)
So, we have:
-3 = a(0 - 1)^2 - 2
Evaluate the like terms
a = -1
Substitute a = -1 in y = a(x - 1)^2 - 2
y = -(x - 1)^2 - 2
Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2
Read more about parabola at:
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Standard form: Ax + By = C
Equation: 12x + 15y = 120
I believe the answer would be B. Catalysts lower the activation energy.
5b-2=13 add 2 to both sides
5b=15 divide both sides by 5
b=3
Answer:
1168/3 or 389 1/3 as a mixed fraction
Step-by-step explanation:
Simplify the following:
4^2 (4×36/6 + 1/3)
4×36/6 = (4×36)/6:
4^2 ((4×36)/6 + 1/3)
4^2 = 16:
16 ((4×36)/6 + 1/3)
36/6 = (6×6)/6 = 6:
16 (4×6 + 1/3)
4×6 = 24:
16 (24 + 1/3)
Put 1/3 + 24 over the common denominator 3. 1/3 + 24 = 1/3 + (3×24)/3:
16 1/3 + (3×24)/3
3×24 = 72:
16 (1/3 + 72/3)
1/3 + 72/3 = (1 + 72)/3:
16 (1 + 72)/3
1 + 72 = 73:
16×73/3
16×73/3 = (16×73)/3:
(16×73)/3
| | 7 | 3
× | | 1 | 6
| 4 | 3 | 8
| 7 | 3 | 0
1 | 1 | 6 | 8:
Answer: 1168/3
