What is the surface area of the box if it is scaled up by a factor of 10? if the surface area is 2 1/2
1 answer:
You might be interested in
Distance of each track are:
D₁ = 428.5 yd
D₂ = 436.35 yd
D₃ = 444.20 yd
D₄ = 452.05 yd
D₅ = 459.91 yd
D₆ = 467.76 yd
D₇ = 475.61 yd
D₈ = 483.47 yd
<u>Explanation:</u>
Given:
Track is divided into 8 lanes.
The length around each track is the two lengths of the rectangle plus the two lengths of the semi-circle with varying diameters.
Thus,
Starting from the innermost edge with a diameter of 60yd.
Each lane is 10/8 = 1.25yd
So, the diameter increases by 2(1.25) = 2.5 yd each lane going outward.
So, the distances are:
D₁ = 240 + π (60) → 428.5yd
D₂ = 240 + π(60 + 2.5) → 436.35 yd
D₃ = 240 + π(60 + 5) → 444.20 yd
D₄ = 240 + π(60 + 7.5) → 452.05 yd
D₅ = 240 + π(60 + 10) → 459.91 yd
D₆ = 24 + π(60 + 12.5) → 467.76 yd
D₇ = 240 + π(60 + 15) → 475.61 yd
D₈ = 240 + π(60 + 17.5) → 483.47 yd