Help me to do this question
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
When Riko left his house, Yuto was 5.25 miles along the path.
Average speed of Riko = 0.35 miles per minute
Average speed of Yuto = 0.25 miles per minute
First we will calculate the time in which Riko will catch Yuto on the track.
Relative velocity of Riko as compared to Yuto will be = velocity of Riko - velocity of Yuto
= 0.35 - 0.25
= 0.10 miles per minute
Now we this relative velocity tells that Riko is moving and Yuto is in static position.
By the formula,
Average speed =
0.10 =
t =
t = 52.5 minutes
Now we know that Rico will catch Yuto in 52.5 minutes. Before this time he will be behind Yuto.
So duration of time in which Rico is behind Yuto will be 0 ≤ t ≤ 52.5
Here time can not be less than zero because time can not be with negative notation. It will always start from 0.
9514 1404 393
Answer:
31.243 units
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the relationships between sides and angles in a right triangle. Using the attached figure, it is convenient to find the length of BE as an intermediate step in the solution.
Sin = Opposite/Hypotenuse
sin(30°) = BE/100
BE = 100·sin(30°)
Then ...
Tan = Opposite/Adjacent
tan(58°) = BE/x
x = BE/tan(58°) = 100·sin(30°)/tan(58°)
x ≈ 31.243 . . . . units
_____
<em>Comment on the figure</em>
The intermediate problem in creating the figure was to locate point D. That was accomplished by locating point C on a line at an angle of 58° CCW from the horizontal, using point B as a center. Then D is the intersection of BC with the x-axis. BE is drawn perpendicular to the x-axis.
