|----------------------------------120 m ------------------------------|
|--------------|--------------|--------------|--------------|--------------|
P Q R
Distance from P to R = 120m
120 ÷ 5 = 24m
The distance from Q to R is 24m.
Step-by-step explanation:
Draw diagonal AC
The triangle ABC has sides 17 and 25
Say AB is 17, BC is 25
Draw altitude on side BC from A , say h
h = 17 sin B
Area = 25*17 sin B = 408
sin B = 24/25
In ∆ ABC
Cos B = +- 7/25
= 625 + 289 — b^2 / 2*25*17
b^2 = 914 — 14*17 = 676
b = 26
h = 17*24/25 = 408/25 = 16.32
Draw the second diagonal BD
In ∆ BCD, draw altitude from D, say DE =h
BD^2 = h^2 + {(25 + sqrt (289 -h^2) }^2
BD^2 = 16.32^2 + (25 + 4.76)^2
= 885.6576 + 266.3424
BD = √ 1152 = 33.94 m
The answer is 0.5 hours.
A velocity (v) of an object is a distance (d) divided by time (t):
v = d ÷ t ⇒ t = d ÷ v
It is given:
v1 = 192 km/h
v2 = 960 km/h
t1 - t2 = 2h
We need to calculate t2.
Since they will travel the same distance before the jet overtakes the plane, we can say that d1 = d2 = d
Now, let's express t1 and t2:
t1 = d/192
t2 = d/960
Therefore: d/192 - d/960 = 2
The least common denominator is 960, so:
5d/960 - d/960 = 2
4d/960 = 2
4d = 2 · 960
4d = 1920
d = 1920 ÷ 4 = 480 km
Now t2 = d/960 = 480/960 = 0.5 hours.
Answer: $21.6
Step-by-step explanation: you multiply 1.20 by 18
