Answer:
56 in
Step-by-step explanation:
Square ....so each side is = x
x * x = area = 196
x^2 = 196
x = sqrt (196) = 14 in
perimeter = x + x + x + x = 56 in
Try this solution:
dx/siny=dy/cos2x;
sinydy=cos2xdx;
cosy=-1/2 sin2x+C.
Answer:
Arc JK: 27°
Arc NJ: 153°
Arc JL: 117°
My guess on arc KNM is 207°
My other guess for MJL is 297
JKM 180°
Step-by-step explanation:
Answer:
Segments perpendicular to the sides of the triangle through the intersection of the angle bisectors were constructed.
Step-by-step explanation:
The above choice represents a bit of excess work. Actually, only one such perpendicular line segment needs to be constructed in order to determine the radius of the inscribed circle.
Once you know the center and radius, you can construct the inscribed circle.
Answer:
24 feet in lenght and 15 in width
Step-by-step explanation:
