You have a fishing line spool with an end that has an area of 20.0 cm2. how much fishing line do you need to wind around the spool 10 times? express your answer (in cm) to three significant figures.

Assuming the spool is a cylinder and the circumference we're winding around is that of a circle with the given area, we can write the relation between circumference and area as

... C = 2√(πA)

10 times the circumference is then

... 10C = 20√(π·20 cm²) = 40√(5π) cm ≈ 159 cm

The usual formulas for circumference and area are

... C = 2πr

... A = πr²

If we multiply the area formula by π and take the square root, we get

... πA = (πr)²

... √(πA) = πr

Multiplying this by 2 gives circumference.

... C = 2√(πA) = 2πr

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°