Answer:
Hey mate, please describe your question perfectly.
Answer:
see below
Step-by-step explanation:
<h3>Given</h3>
- Distance is 142.2 m, correct to 1 decimal place
- Time is 7 seconds, correct to nearest second
<h3>To find:</h3>
- Upper bound for the speed
<h3>Solution </h3>
<em>Upper bound for the speed = upper bound for distance/lower bound for time</em>
- Upper bound for distance = 142.25 m (added 0.1/5 = 0.05)
- Lower bound for time = 6.5 seconds (subtracted 1/2 = 0.5)
<u>Then, the speed is:</u>
- 142.25/6.5 = 21.88 m/s
- 21.88 = 21.9 m/s correct to 1 decimal place
- 21.88 = 22 m/s correct to nearest m/s
OM=18, so OQ=QM=18/2=9.
Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
From given statements of congruent NT and OU will also be congruent or identical. So, NT=OU=13
Answer:
y=2x
Step-by-step explanation:
rise/run 2/1
