The speed of wind and speed of plane in still air are 23 and 135
km/h respectively.
<u>Step-by-step explanation:</u>
Let the speed of wind and speed of plane in still air are w and p km/h respectively.
The effective speed on onward journey was
................(1)
The effective speed on return journey was
..............(2)
Adding equation (1) and equation (2) we get,
⇒ ![(p+w)+(p-w) = 158+112](https://tex.z-dn.net/?f=%28p%2Bw%29%2B%28p-w%29%20%3D%20158%2B112)
⇒ ![2p = 270](https://tex.z-dn.net/?f=2p%20%3D%20270)
⇒ ![p = 135](https://tex.z-dn.net/?f=p%20%3D%20135)
Putting value of
in
we get:
⇒ ![p+w=158](https://tex.z-dn.net/?f=p%2Bw%3D158)
⇒ ![135+w=158](https://tex.z-dn.net/?f=135%2Bw%3D158)
⇒ ![w=23](https://tex.z-dn.net/?f=w%3D23)
Therefore ,The speed of wind and speed of plane in still air are 23 and 135
km/h respectively.
Answer:
A = 113.1 in^2
Step-by-step explanation:
The diameter is 12 inches
The radius is d/2 = 12/2 = 6 inches
The area of a circle is
A = pi r^2
Assuming pi is used by the pi button on the calculator
A = pi * 6^2
A =113.097335
Rounding to the nearest tenth
A = 113.1
If the length of the trail on map A is 9 centimeters, and the scale factor is a centimeter to 2 kilometers, the actual length of the trail would be 18 kilometers.
Answer:
11. 90 degrees...right angle
12. 120 degrees...obtuse angle
13. 30 degrees...acute angle
Step-by-step explanation:
Well, area is Length•Width. So 6•8=48 I’ve got a pretty good feeling! Hope this helps!