Answer:
B and C
Step-by-step explanation:
![56 - ( {3}^{3} + 13) = 16](https://tex.z-dn.net/?f=56%20-%20%28%20%7B3%7D%5E%7B3%7D%20%20%2B%2013%29%20%3D%2016)
For A
![\frac{92 - (12 + 16)}{4} = 16 \\](https://tex.z-dn.net/?f=%20%5Cfrac%7B92%20-%20%2812%20%2B%2016%29%7D%7B4%7D%20%20%3D%2016%20%5C%5C%20)
For B
![{6}^{2} \div 4 + 5 = 14](https://tex.z-dn.net/?f=%20%7B6%7D%5E%7B2%7D%20%20%5Cdiv%204%20%2B%205%20%3D%2014)
For C
![(6 \times 7) + 3 + 2 = 47](https://tex.z-dn.net/?f=%286%20%5Ctimes%207%29%20%2B%203%20%2B%202%20%3D%2047)
For D
![19 \times 4 - {8}^{2} + 4 = 16](https://tex.z-dn.net/?f=19%20%5Ctimes%204%20-%20%20%7B8%7D%5E%7B2%7D%20%20%2B%204%20%3D%2016)
Here, B and C have not same value
18+18=36. 20% is 9 apples. 36+9=45. 45 apples at the stand. (not good for business)
Point R is the midpoint of FE¯¯¯¯¯ , so the coordinates of point R are (3a, b).
In △DEF , the length of the base, DF¯¯¯¯¯ , is
4a, and the height is 2b, so its area is
1/2×4a×2b = 4ab.
In △QRP , the length of the base, QR¯¯¯¯¯ , is
3a-a = 2a, and the height is b, so its area is 1/2×2a×b = ab .
Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the larger isosceles triangle.
I believe the answer to the question is D.