Answer:
In the figure ∠ABO and ∠BCO have measures equal to 35°.
Step-by-step explanation:
Measure of arc AD = 180-measure of arc CD= 180-125 =55
m<AOB= 55 ( measure of central angle is equal to intercepted arc)
<OAB= 90 degrees (Tangent makes an angle of 90 degrees with the radius)
In triangle AOB ,
< AB0 = 180-(90+55)= 35 degrees( angle sum property of triangle)
In triange BOC ,< BOC=125 ,
m<, BCO=35 degrees
Hello from MrBillDoesMath!
Answer:
12/5
Discussion:
Note that 10 is a common denominator of both denominators:
9/5 = (9*2)/(5*2) = 18/10
- ( -6/10) = + (6/10)
So the original problem is equivalent to
18/10 + 6/10 =
(18 +6)/10 =
24/10 =
(2*12)/ (2*5) => cancelling the common factor "2"
12/5 =
2 and 2 fifths =
22/5 (though this looks like (22)/5!)
Thank you,
MrB
2929293939393939494. merry iwie i
The time needed to run the second half of the race is 8.44 s
Step-by-step explanation:
The total distance covered by the athlete during the race is
d = 200 m
And the total time taken to cover this distance is
T = 19.19 s
We also know that the time the athlete needs to cover the first half of the race is
Also, we know that
where 
is the time the athlete takes to cover the second half of the race.
Re-arranging this equation and susbtituting the values, we find the value of t2:
Learn more about uniform motion:
brainly.com/question/8893949
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