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Elden [556K]
3 years ago
14

Find the area of a square with side length 4/5 yard.

Mathematics
2 answers:
melisa1 [442]3 years ago
8 0

Answer:

Step-by-step explanation:

The area of a square is the side squared.

A=s^2, if s=4/5 then

A=(4/5)^2

A=16/25

So the area is 16/25 yd^2 as a fraction or 0.64 yd^2 as a decimal.

BigorU [14]3 years ago
6 0

Answer:

side length = 0.8 yards

area = side length^2= 0.8^2 = 0.64 or 64/100

Step-by-step explanation:

hope that helped

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Two bicyclists are 7/8 of the way through a mile long tunnel when a train approaches the closer end at 40 mph. The riders take o
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Answer:

the cyclists rode at 35 mph

Step-by-step explanation:

Assuming that the cyclists stopped, and accelerated instantaneously at the same speed than before but in opposite direction , then

distance= speed*time

since the cyclists and the train reaches the end of the tunnel at the same time and denoting L as the length of the tunnel :

time = distance covered by cyclists / speed of cyclists = distance covered by train / speed of the train

thus denoting v as the speed of the cyclists :

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v = 7/8 * 40 mph = 35 mph

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3 years ago
What is the measurement of ∠G? Show your work.
san4es73 [151]

Answer:

x = 40

Step-by-step explanation:

The angles add to 180 degrees

30+110 +x = 180

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140+x-140 =180-140

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3 years ago
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57%:43% of 3.09 million
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Answer:

Step-by-step explanation:

Percentage Calculator: What is 57 percent of 3.09? = 1.7613.

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2 years ago
A person on a runway sees a plane approaching. The angle of elevation from the runway to the plane is 11.1° . The altitude of th
Gnoma [55]

Answer:

The horizontal distance from the plane to the person on the runway is 20408.16 ft.

Step-by-step explanation:

Consider the figure below,

Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner.  The angle of elevation from the runway to the plane is 11.1°

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\tan\theta=\frac{Perpendicular}{base}

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\tan\theta=\frac{AB}{BC}

\tan 11.1^{\circ} =\frac{4000}{BC}

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BC=\frac{4000}{\tan 11.1^{\circ} }

BC=\frac{4000}{0.196} (approx)

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Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft

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