Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by

The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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Answer:
option 4. The graph has zeroes at x = 15 an x = -15 and it opens upward.
Step-by-step explanation:
we have

This is the equation of a vertical parabola open upward (the leading coefficient is positive)
The vertex represent a minimum
The vertex is the point (0,-225)
The axis of symmetry is x=0 (y-axis)
Find the x-intercepts (values of x when the value of y is equal to zero)
For y=0

square root both sides

therefore
The graph has zeroes at x = 15 an x = -15 and it opens upward.
H= volume/ pi(r)^2 is the height of a cylinder
406 divided by 140 makes 2.9 multiply that 100 and you arrive at 290%. Hope this helps!
Answer:
where are the followings
Step-by-step explanation:
give the following .
maybe I can help you.
thankyou