In this case, we cannot simply take the average speed by
adding the two speeds and divide by two.
What we have to do is to calculate the time required
going to school and the return trip home.
We know that to calculate time, we use the formula:
t = d / v
where,
d = distance = 4.8 km = 4800 m
v = velocity
Let us say that the variables related to the trip going
to school is associated with 1, and the return trip home is 2. So,
t1 = 4800 m / (22.6 m / s)
t1 = 212.39 s
t2 = 4800 / (16.8 m / s)
t2 = 285.71 s
total time, t = t1 + t2
t = 498.1 s
Therefore the total average velocity is:
= (4800 m + 4800 m) / 498.1 s
= 19.27 m / s = 19.3 m / s
Answer:
19.3 m/s
Answer:
The ratios are both identical. (4 over 5 and 4 over 5)
Step-by-step explanation:
Th picture below represent the image you are referring.
sin x = opposite / hypotenuse
opposite side = 4
hypotenuse = 5
sin x° = 4/5
cos y° = adjacent/hypotenuse
adjacent = 4
hypotenuse = 5
cos y° = 4/5
The ratios are both identical (4 over 5 and 4 over 5)
(2x² - 4x + 3) + (x² - 4x - 4) =
2x² - 4x + 3 + x² - 4x - 4 =
3x² - 8x -1
B.
