Answer:
see image...
start with Graphing Form: y = a(x-h)²+ k
knowing that h is the x value (negative of it) of the vertex , and k is the y value of the vertex
Graphing Form Equation: y = a(x+3)²+32
you are also given that when the x value is 5 the y value has to be 0
0 = a(4)^2 + 32
a has to be - 1/2
Step-by-step explanation:
Answer:
3rd answer down
Step-by-step explanation:
ok so you take the median of all the #'s which turns out to be 126.5 and that is where your bar should point to in the box, then you should start with 105, and your last bit of the line should end at 150, the start of your box should start at the median of 105 and 117, and the end of your box should stop at the median of 145 and 150.
Option C: There are no solutions
Explanation:
The linear equations is graphed.
We need to determine the solution of the system of equations.
The solution of the equations can be determined by finding the point of intersection of the two equations.
From the figure, it is obvious that the two equations are parallel to each other.
Also, the parallel lines have the same slope and the parallel lines never intersect.
Hence, the system consisting of parallel lines have no solution.
Therefore, the solution to the system of linear equations graphed is no solution.
Thus, Option C is the correct answer.
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Let the area be y .
Area = (base) × (height)
Base = 2x
Height = h
Let the area of the rectangular pens be y .
∴ y = 2xh
Perimeter of all the fencing = 4x+3h
∴ 4x+3h = 120
now we solve for h
3h = 120-4x
h = 40 - 4/3 x
Now we will substitute this value in the above first equation:
y = 2xh
or, y = 2x (40 - 4/3 x)
or, y = 80x - 8/3 x²
Now for the maximum area we have to find the first order differentiation of y
now,
dy /dx = 80 - 16/3 x
At dy/dx = 0 we get the value of x for which y is maximum.
80 - 16/3 x = 0
or, - 16/3 x = -80
or, x = 15 feet
Hence height = 40 - 4/3 x = 40 - 20 = 20feet
Maximum area = 2xh = 2×15×40 = 1200 square feet
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Disclaimer : The missing figure for the question is attached below.
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brainly.com/question/27531272
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