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Tresset [83]
2 years ago
14

An airplane flies to San Francisco from Los Angeles in 4 hours. It flies back in 3 hours. If the wind is blowing from the north

at a velocity of 20 mph during both flights, what was the airspeed of the plane (its speed in still air)?
Mathematics
1 answer:
Sedaia [141]2 years ago
4 0

Answer:

Airspeed in still air:  140 mph

Step-by-step explanation:

Let s represent the speed in still air.  Recognize that the distance traveled is the same in either direction.

Also note that SF is north of LA.

Distance to SF from LA     =      Distance to LA from SF

(s-20)(mph)(4 hr)                 =    (s+20)(mph)(3 hr)

Then:

4s - 80 = 3s + 60 => s = 60 + 80 + 140 (mph)

The airspeed of the plane was 140 mph in still air.  With a tail wind, the plane  is faster; with a headwind, slower.

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Help please ill give brainliest answer
mihalych1998 [28]

Answer:

ΔLMN ≅ ΔLQP by (SAA)

Step-by-step explanation:

It is given that line (NM) is congruent to the line (PQ), meaning they have the same measure. This is signified by the small red line on each of these sides.

Moreover, it is also given that angle (MNL) is congruent to angle (QPL), this is shown by the red arc around these angles.

Finally one can figure out that angle (NLM) is congruent to angle (PLQ) by the vertical angles theorem. The verticle angles theorem states that when two lines intersect, the opposite angles are congruent.

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6 0
2 years ago
A water storage tank has the shape of a cylinder with diameter 14 ft. It is mounted so that the circular cross-sections are vert
andrezito [222]

Answer:

percentage of the total capacity is 75.6%

Step-by-step explanation:

Hello! To solve this problem we follow the following steps

1. draw the complete scheme of the problem (see attached image)

2. To solve this problem we must find the area of ​​the circular sector using the following equation.(c in the second attached image)

A=\frac{R^2}{2} (\alpha -sin\alpha )

\alpha =2arccos(\frac{d}{R})

3. observing the attached images we replace the values ​​in the equations and find the area of ​​the circular sector, remember that you must transform the angle to radians

\alpha =2arccos(\frac{5}{7})=88.83

A=\frac{R^2}{2} (\alpha -sin\alpha )\\A=\frac{7^2}{2} (88.83-sen88.83)*\frac{\pi rad}{180} =37.57ft^2

4.we calculate the area of ​​the total circle (At), then subtract the area of ​​the circular sector (Ac) to find the area occupied by water (Aw)

At=\frac{\pi }{4} (14ft)^2=153.93ft^2

Aw=At-Ac=153.93-37.57=116.36ft^2

5.Finally, we calculate the percentage that represents the water in the tank by dividing the area of ​​the water over the total area of ​​the tank

\frac{116.36}{153.93} *100=75.6

percentage of the total capacity is 75.6%

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