If m varies directly with x, then that means that you have to find the slope. To make it easier, m = y-variable, and x = x-variable.
Because you are trying to find your constant of variation (another word for slope, rate of change, etc.), the change of y over the change of x is your equation. in other words, divide your m variable by x.
When m = 60 and x = 6, you can set it up. 60 / 6 = k; and k = 10 when solved. Since 10 is not one of the answer choices listed in your answer key, I believe that your answer is NONE OF THE ABOVE.
I hoped this helped you.
Answer:
an Injective function is a function that maps distinct elements of its domain to distinct elements of its codomain. In other words, every element of the function's codomain is the image of at most one element of its domain.
Step-by-step explanation:
Answer:
$123,200
Step-by-step explanation:
0.06 x 2 = 0.12
140,000 x 0.12 = 16,800
140,000 - 16,800 = 123,200
1/3 + 5/6 + 5/12
1.583333333333333
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:
brainly.com/question/1480401
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