I can’t see the pictures if there are any
Since, the polygon is a trapezoid made up of a rectangle and a right triangle. Therefore, according to the question, the figure of the polygon is attached.
Since, perimeter is the total length of the outer boundary of the figure. Therefore,
Perimeter of the polygon is
Area of the polygon = Area of Rectangle + Area of Triangle
![=[(18) \times (15)] + [(\frac{1}{2}) \times (8) \times (15)]](https://tex.z-dn.net/?f=%3D%5B%2818%29%20%5Ctimes%20%2815%29%5D%20%2B%20%5B%28%5Cfrac%7B1%7D%7B2%7D%29%20%5Ctimes%20%288%29%20%5Ctimes%20%2815%29%5D)
![=270 + [(\frac{8}{2}) \times (15)]](https://tex.z-dn.net/?f=%3D270%20%2B%20%5B%28%5Cfrac%7B8%7D%7B2%7D%29%20%5Ctimes%20%2815%29%5D)
![=270 + [4 \times (15)]](https://tex.z-dn.net/?f=%3D270%20%2B%20%5B4%20%5Ctimes%20%2815%29%5D)
Answer:
Next 3 terms 031, N30, D31
Step-by-step explanation:
J30, J31, A31, S30
The above series represent the
first letter of the month and the days in it
J30= June which have 30 days
J31= July which has 31 days
A31= August which have 31 days
S30= September which have 30 days
30 days hath September, April, June and November. All the rest have 31 except February alone which has but 28 or 29 days
The next 3 terms of the series J30, J31, A31, S30 are 031, N30, D31
Use the law of cosines.
a2+b2−2abcosC=c2
Find the measure of angle C. It is the opposite side of c.
c2−a2−b2−2ab=cosC
cosC=13.62−22.52−182−2(22.5)(18)≈0.797
C=cos−10.797=0.649 rad=37.19∘
angle B:
a2+c2−2accosB=b2
cosB=b2−a2−c2−2ac
B=cos−1b2−a2−c2−2ac=cos−1182−22.52−13.62−2(22.5)(13.6)≈0.927 rad=53.13∘
angle A:
b2+c2−2bccosA=a2
A=89.68∘
24 x 1/2 = 12,
use these numbers to make a problem
