Answer:
Option A. one rectangle and two triangles
Option E. one triangle and one trapezoid
Step-by-step explanation:
step 1
we know that
The area of the polygon can be decomposed into one rectangle and two triangles
see the attached figure N 1
therefore
Te area of the composite figure is equal to the area of one rectangle plus the area of two triangles
so
![A=(8)(4)+2[\frac{1}{2}((8)(4)]=32+32=64\ yd^2](https://tex.z-dn.net/?f=A%3D%288%29%284%29%2B2%5B%5Cfrac%7B1%7D%7B2%7D%28%288%29%284%29%5D%3D32%2B32%3D64%5C%20yd%5E2)
step 2
we know that
The area of the polygon can be decomposed into one triangle and one trapezoid
see the attached figure N 2
therefore
Te area of the composite figure is equal to the area of one triangle plus the area of one trapezoid
so
Answer:
x° = ∠OBR = ∠ABC (base angles of a cyclic isosceles trapezoid)
Step-by-step explanation:
APRB form a cyclic trapezoid
∠APO = x° (Base angle of an isosceles triangle)
∠OPR = ∠ORP (Base angle of an isosceles triangle)
∠ORB = ∠OBR (Base angle of an isosceles triangle)
∠APO + ∠OPR + ∠OBR = 180° (Sum of opposite angles in a cyclic quadrilateral)
Similarly;
∠ORB + ∠ORP + x° = 180°
Since ∠APO = x° ∠ORB = ∠OBR and ∠OPR = ∠ORP we put
We also have;
∠OPR = ∠AOP = ∠BOR (Alternate interior angles of parallel lines)
Hence 2·x° + ∠AOP = 180° (Sum of angles in a triangle) = 2·∠OBR + ∠BOR
Therefore, 2·x° = 2·∠OBR, x° = ∠OBR = ∠ABC.
Answer:
A
x=0.625y
y=1.6x
B
y=3.2
Step-by-step explanation:
Answer:
Step-by-step explanation:
This is a good example of permutations. For the first slot on his shelf, Dylan has 7 trophies to choose from. Then 6, 5, and so on. Since he is only fitting 4 trophies on his shelf, we only continue until we have the slots filled. Thus, we have 
to do so.
*Note: We do not divide by 
because in this case, order matters. A different order of the same four books still produces different arrangements, so the order matters.
