D midpoint of EC -----------------> FD parallel to AC and FD=AC/2=14/2=7
<span>2-EB=EA E midpoint of AB </span>
<span>DB=DC D midpoint BC ...............> ED=AC/2=2 </span>
<span>3-T midpoint of SR </span>
<span>U midpoint of QR ---------> TU = QS/2 </span>
<span>QS=2 TU = 4.4 </span>
<span>4- The same steps SR=2 UV=9 </span>
<span>5-N midpoint of KM </span>
<span>O midpoint of ML </span>
<span>* NO parallel to Kl</span>
Answer:
87.92 inches cubed or 87.92 in.^3
Step-by-step explanation:
The base of a cylinder is a circle, so we need to calculate the area of the circle first. The equation for the area of a circle is:
area (a) = pi * r^2 =3.14*2*2 =12.56sq.inches
again,
volume of cylinder = area of the base * the height
= 12.56 * 7
= 87.92 inches cubed or 87.92 in.^3
We call x the 4 points-worth problems and y the 3 points-worth problems
You know that x+y = 32
4x+3y = 111
You know that the difference from the x and y is 32 so write:
x = 32-y
Substitute at x the value of 32-y
4(32-y)+3y = 111
128-4y +3y = 111
-4y+3y = 111-128
-y = -17
y = 17
Answer:
Step-by-step explanation:
Let the other side of the rectangle be y. The perimeter of the rectangle is expressed as P = 2(x+y)
Given P = 30ft, on substituting P = 30 into the expression;
30 = 2(x+y)
x+y = 15
y = 15-x
Also since the area of the rectangle is xy;
A = xy
Substitute y = 15-x into the area;
A = x(15-x)
A = 15x-x²
The function that models its area A in terms of the length x of one of its sides is A = 15x-x²
The side of length x yields the greatest area when dA/dx = 0
dA/dx = 15-2x
15-2x = 0
-2x = -15
x = -15/-2
x = 7.5 ft
Hence the side length, x that yields the greatest area is 7.5ft.
Since y = 15-x
y = 15-7.5
y = 7.5
Area of the rectangle = 7.5*7.5
Area of the rectangle = 56.25ft²
Answer: IM pretty sure your correct
Step-by-step explanation:
