Answer:
Step-by-step explanation:
A
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:2
Step-by-step explanation:
Slope = rise/run = (8-2)/(5-2) = 6/3 = 2
The answer is 3 because if u divide 81 and 1/3 3 times it makes 3. So there u go <span />
For this, we need to know the length of the base at the time of interest. It will be
... A = (1/2)bh
... b = (2A)/h = 2(81 cm²)/(10.5 cm) = 108/7 cm
Differentiate the formula for area and plug in the given numbers.
... A = (1/2)bh
... A' = (1/2)(b'h +bh')
... 3.5 cm²/min = (1/2)(b'·(10.5 cm) + (108/7 cm)·(2.5 cm/min))
... 7 cm²/min = 10.5b' cm + 38 4/7 cm²/min . . . . simplify a bit
... -31 3/7 cm²/min = 10.5b' cm . . . . . . . . . . . . . . . subtract 38 4/7 cm²/min
... (-220/7 cm²/min)/(10.5 cm) = b' ≈ -3.0068 cm/min
The base is changing at about -3 cm/min.