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Kazeer [188]
4 years ago
9

A parachutist's speed during a free fall reaches 70 meters per second. What is this speed in feet per second? At this speed, how

many feet will the parachutist fall during 20 seconds of free fall? In your computations, assume that 1meter is equal to 3.3 feet
Mathematics
1 answer:
andrew11 [14]4 years ago
5 0

Speed is 231 ft/s.

Distance is 4620 ft.

Solution:

First we have to convert the unit of speed from m/s to ft/ s.

Then for 20 seconds, we have to calculate the feet covered by the parachute.

1 meter $=3.3$ feet\\\\$\frac{70 \frac{m}{s} \times 3.3 f t}{1 m}=231 \frac{f t}{s}$

In 20 seconds, the parachute covers the distance of,

231 \frac{f t}{s} \times 20 s=4620 \mathrm {ft}

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Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

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Let,

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BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

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Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

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Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

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DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

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