Recliner:
Markup $10
selling price $300
computer:
markup $19.20
selling price $67.20
The answer would be C. We know that d is equal to the initial depth of a lake. The two given initial depths are 58 feet and 53 feet, so we know that one of the equations must be either d=58 or d=53. Because there only C has either one of those, d=58, we know that it must be the answer.
To find the other equation, it is just a linear function for the other lake. The y-intercept, or initial value, is 53, so in the equation y=mx+b, it is the b value. The slope, or m value, is 3 feet, so you have y=d=3x+53.
Answer:
5
Step-by-step explanation:
I hope this helps you
3 1/8=3.8+1/8=25/8
25/8-4/7
25.7/8.7-4.8/7.8
175-32/56
143/56
2 31/56
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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