Answer:
Step-by-step explanation:
1)
O = center of circle (origin)
we know that the angle at the center of the circle ∠ ROS will be
180= 2(31) + ∠x
180-62 = ∠x
118° = ∠x
The supplemental angle to the 118° will be 62°
62° is the interior angle to arc QR , so
arc QR is also 62°
3)
b/c the intercepted arc YZ = 2* 68=136
then 136+125+? = 360
? = 99°
arc ZX = 99°
5)
O= center point
we are given the two arcs 120 and 70 for both of these we know that the
interior angles will be the same. ∠JOX has a central angle of 120 , then
b/c triangle JOX is an isosceles, we know that the two angles J and X of
the triangle JOX will be 1/2 of 60 , or 30 each
also for
also ∠XOY has an interior angle of 70 so the two angle at X and Y will be 1/2 of 110 , or 55°
now add 55+30 to find X for box XYZJ
85° = ∠X
Answer:
The Answer is D - $34.19
Step-by-step explanation:
Have a nice day.
Answer:
Principal=₹42800
Rate of interest=5% but 2 years so multiply by 2 =5×2=10%
Time=2 years
Simple interest=42800×10×2/100= ₹8560
Amount=42800+8560=Rs51360
The answer is ₹ 51360
Please make me as brainliest
Step-by-step explanation:
The midpoint of a line divides the line into equal segments.
The option that proves PQ = LO is (a)
The given parameters are:




P is the midpoint of LM.
So, we have:



Q is the midpoint of NO.
So, we have:



Distance PQ is calculated as follows:

This gives:



Distance LO is calculated as follows:



So, we have:


Thus:

Hence, the correct option is (a)
Read more about distance and midpoints at:
brainly.com/question/11231122
Answer:
b=5..............is the answer