Answer:
A racer has just completed his first lap at a speed of 198m/s. Now his speed decrease or he decelerateted at 195m/s after 2s of his completion of first lap, as soon as he saw a curve path(line segment R). He then moved at a constant speed of 195m/s for the next 2s(line segment s). Then the race track after 4s of the completion of first lap was straight, so the racer accelerated to 200m/s within 2s( line segment T) and traveled at a constant speed for the next 6s( line segment U). After 12s from the first lap completion, he immediately saw a curve track, so he decelerateted to the speed of 194m/s within 3s( line segment V).
Hope it helped.
please mark brainliest
Well to be honest, you need to straight forward to her and tell her that you do not appreciate her copying you and tell her you dont want to be rude or anything, but this needs to stop.
Answer:
The actual speed of the car is either 57 mph or 63 mph
Step-by-step explanation:
Given as :
The measured speed o the car = 60 mph
The percentage error in speedometer = 5 %
Let The actual speed of car = x mph
Now, percentage error = 
× 100
So, 
= 
or, × 60 = x - 60
Or, 3 = x - 60
∴ x = 63 mph
Hence The actual speed of the car is either 57 mph or 63 mph . Answer
Answer:
23,4
Step-by-step explanation:
<u><em>hi there user! Looks like your looking for help, so Mizuki came here to help you!</em></u>
the answer is:
<em>m is 2/3</em>
The whole work I did:
m + 4 ÷ 3 = 2
m + 4/3 = 2
m + 4/3 - 4/3 = 2 - 4/3
m = 6/3 - 4/3
m = 2/3
Anyways hope this helped! qwq
