We know that
∠ TSR = 84°
if SQ bisects ∠ <span>TSR
then
</span>∠ RSQ = ∠ TSR/2
<span>so
</span>∠ RSQ = (1/2)*84°----- 42°
∠ RSQ = 3x-9
3x-9=42-------> 3x=42+9------> 3x=51-----> x=51/3-----> x=17°
<span>
the answer is
</span>x=17°<span>
</span>
Answer:
4 rows
Step-by-step explanation:
GCF of 12 and 16 is4
Answer:
haha pues cuesta mas las hamburgesas porque toca ir al doctor despues
Step-by-step explanation:
Answer:
13.98 in²
Step-by-step explanation:
I don't understand it, either.
Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which <u>is not</u> an answer choice.
__
The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which <u>is</u> an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.
The area of a segment is given by the formula ...
A = (1/2)(θ -sin(θ))r² . . . . . . where θ is the central angle in radians.
Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...
A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²
Rounded to hundredths, this is ...
≈ 13.98 in²
