At the beginning it was 22 inches below normal (-22).
Then it decreased by 3 1/6 (-19/6).
Then increased by 1 6/7 (+13/7)
-22 - 19/6 + 13/7
-924/42 - 133/42 + 78/42
-1057/42 + 78/42
-979/42
-23 13/42
It was 23 13/42 inches below normal after April.
Using the relation between velocity, distance and time, it is found that:
- The rate of the plane in still air is of 140 mph.
- The rate of the wind is of 40 mph.
<h3>What is the
relation between velocity, distance and time?</h3>
Velocity is <u>distance divided by time</u>, that is:
A plane traveling with the wind travels 900 miles in 5 hours, hence:
The return trip is against the wind and takes 9 hours, hence:
From the first equation, we have that:
![v_w = 180 - v_a{/tex]Replacing on the second:[tex]v_a - v_w = 100](https://tex.z-dn.net/?f=v_w%20%3D%20180%20-%20v_a%7B%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EReplacing%20%3C%2Fstrong%3Eon%20the%20second%3A%3C%2Fp%3E%3Cp%3E%5Btex%5Dv_a%20-%20v_w%20%3D%20100)
[tex]v_w = 180 - 140 = 40{/tex]
Hence:
- The rate of the plane in still air is of 140 mph.
- The rate of the wind is of 40 mph.
More can be learned about the relation between velocity, distance and time at brainly.com/question/24316569
You have to include a drawing that relates the distace between de towers and some angles.
I will use one that gives the angle from the base of Seafirst Tower to the top of Columbia tower as 53 degress.
This lets you calculate the distance between the towers, d, as
tan(53) = 954 / d => d = 954 / tan(53) = 718.89ft
The same drawing gives the angle from the the base of the Columbtia tower to the top of the Seafirst Tower as 27 degrees.
Tnen, tan(27) = height / d => height = d*tan(27) = 718.89*tan(27) = 366.29 ft
Answer: 366.29 ft
Answer:
a=14
b=2
c=3
Step-by-step explanation:
a) 2+ab= 2+(3*4)
= 2+12
= 14
Answer:
Step-by-step explanation:
a=-2.9
b=-2 5/8=-21/8=-2.625
c=-7/3=-2.333333
