A bicycle wheel with diameter 16 inches rides over a screw in the street. The screw is on level ground before it punctures the bike's tire. After the bike has moved forward another 56 inches, how high above the ground is the
screw? Round to the nearest tenth of an inch.
O 4 inches
O 8 inches
O 12 inches
O 16 inches

16 inches

Step-by-step explanation:

Note: it should say 56π inches instead of 56 inches otherwise no option is correct

Given

Full circumference of the wheel

Bike moves

The wheel turns:

We only consider her 0.5 of full turn of the wheel to identify the screw position

Since we consider the turns from the ground level as the initial position of the screw, the screw position in the end will be half turn opposite to ground level.

This is the full diameter length, which is 16 inches

So correct answer option is 16 inches

i just took the test, the correct answer is .. 31.5 in.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°