<span>To minimize the perimeter you should always have a square.
sqrt(289) = 17
The dimensions should be 17 X 17
To see , try starting at length 1, and gradually increase the length.
The height decreases at a faster rate than the length increases, up until you reach a square.
Or if you want to use algebra, Say the width is 17-x
Then the length is 289/(17-x)
Now, this is bigger than 17+x, as shown here:
289/(17-x) > 17+x
289 > 289 - x^2
which is true.
so the perimeter would be bigger than 2 * (17- x + 17 + x) = 2 * (2 * 17) = 4 * 17
Again, the dimensions should be a square. 17 X 17.</span>
The positive trend among the homework scores and the test scores would reveal that if a student does well in homework, he is likely to do well in exams. This can be explained in the sense that the homework will serve as the practice for the student before finally taking the exam.
Answer:
see below
Step-by-step explanation:
f(x) = −16x^2 + 24x + 16
Set equal to zero to find the x intercepts
0 = −16x^2 + 24x + 16
Factor out -8
0 = -8(2x^2 -3x-2)
Factor
0 = -8(2x +1) (x-2)
Using the zero product property
2x+1 =0 x-2 =0
x = -1/2 x=2
The x intercepts are -1/2 ,2
Since the coefficient of x^2 is negative the graph will open down and the vertex will be a maximum
The x value of the maximum is 1/2 way between the zeros
(-1/2+2) /2 = 1.5/2 =.75
To find the y value substitute into the function
f(.75) = -8(2x +1) (x-2)
=-8(2*.75+1) (.75-2)
= -8(2.5)(-1.25)
=25
The vertex is at (.75, 25)
We have the zeros, and the vertex. We know the graph is symmetrical about the vertex
R = 6t.....subbing in (8,48).....t = 8 and r = 48
48 = 6(8)
48 = 48 (correct)
r = 6t...subbing in (13,78)...t = 13 and r = 78
78 = 6(13)
78 = 78 (correct)
so u have 2 sets of points on this line and they are (8,48) and (13,78)
