Answer:ññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññ
Answer:
Y TO THE POWER 3 BY DIVIDING Y ON BOTH SIDES YOU WILL GET Y TO THE POWER 3
Step-by-step explanation:
Answer:
The parent function is y = x³.
After a vertical stretch by a factor of 3, obtain
y = 3x³
After a horizontal shift 4 unit to the right, obtain
y = 3(x - 4)³
After a vertical shift 3 units down, obtain
y = 3(x - 4)³ - 3
Answer: y = 3(x - 4)³ - 3
The formula for distance is equal to:
d = v * t
where d is distance, v is velocity or speed, and t is
time
Since the distance travelled by the two airplane is
similar, therefore we can create the initial equation:
v1 * t1 = v2 * t2
We know that v1 = 496, and v2 = 558 so:
496 t1 = 558 t2
or
x = 558 t2 / 496
We also know that airplane 1 travelled 30 minutes (0.5
hours) earlier than airplane 2, therefore:
x = t2 + 0.5
Hence,
496 (t2 + 0.5) = 558 t2
496 t2 + 248 = 558 t2
t2 = 4 hours
x = t2 + 0.5 = 4 + 0.5
x = 4.5 hours
So the equation is:
x = 558 t2 / 496
And the first plane travelled:
x = 4.5 hours
