Answer:
The time of a commercial airplane is 280 minutes
Step-by-step explanation:
Let
x -----> the speed of a commercial airplane
y ----> the speed of a jet plane
t -----> the time that a jet airplane takes from Vancouver to Regina
we know that
The speed is equal to divide the distance by the time
y=2x ----> equation A
<u><em>The speed of a commercial airplane is equal to</em></u>
x=1,730/(t+140) ----> equation B
<u><em>The speed of a jet airplane is equal to</em></u>
y=1,730/t -----> equation C
substitute equation B and equation C in equation A
1,730/t=2(1,730/(t+140))
Solve for t
1/t=(2/(t+140))
t+140=2t
2t-t=140
t=140 minutes
therefore
The time of a commercial airplane is
t+140=140+140=280 minutes
Answer:
74.8 m
Step-by-step explanation:
We are given that
Speed of peacock,v=10 km/h
Time,t=1 min 12 sec=60+12=72 s=
1 hour=3600 s
Distance=vt=
Using 1 km=1000 m
We know that
Using the formula
Hence, the peacock was at height of 74.8 m on the tree.
5x is a short term for 5x X, so u just need to do 30/5 and that is what x equals.
When a population or group of something is declining, and the amount that decreases is proportional to the size of the population.
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