I saw the image that should have been posted with this word problem.
The rolls of wrapping paper is cylindrical in shape.
Roll A : length = 30 inches ; diameter = 3 inches
Roll B : length = 24 inches ; diameter = unknown
Lateral surface area = 2 π r h
Lateral Surface area of Roll A = Lateral Surface area of Roll B
2 * 3.14 * 1.5in * 30in = 2 * 3.14 * r * 24 in
282.60 in² = 150.72 in * r
282.60 in² / 150.72 in = r
1.875 in = r
diameter is 2 times the radius = 1.875 in * 2 = 3.75 inches
The diameter of Roll B is 3.75 inches.
1. Place your compass point on X and measure the distance to point Y. Swing an arc of this size above (or below) the segment.
2. Without changing the span on the compass, place the compass point on Y and swing the same arc, intersecting with the first arc.
3. Label the point of intersection as the third vertex of the equilateral triangle.
4. Use you straight edge and connect the points.
5 cent = 0.05x
<span><span>2 cent </span>= 0.02(x+50)</span>
4 cent = 0.04(2x-10)
0.05x + 0.02(x+50) + 0.04(2x-10)
0.05x + 0.02x +1 + 0.08x -0.4 = 4.35
0.15x +0.6 = 4.35
0.15x = 3.75
X = 3.75 / 0.15 = 25
<span> 25 5 cent stamps (
25*0.05 = 1.25)</span>
25+50 = 75 2 cent stamps ( 75*0.02 = 1.50)
2*25 =50-10 = 40 4 cent stamps ( 40*0.04 =1.60)
1.25 + 1.50 +1.60 = 4.35
5 cent = 25
2 cent = 75
4 cent = 40
Answer:
64
Step-by-step explanation:
