Answer:
based on the given formula the amount calculated is for the beginning of the year.
substituting 1 you get 4500 which is beginning of the year 1 when you deposited.
based on that logic substituting 7 for beginning of year 7 will be
4500 + 6 * 0.02 * 4500
4500 + 540
5040
option A.
Establish two right triangles, both with the height of the pole, h.
Call x the distance from the pole to one stake. Then the distance from the other stake to the pole is 6 -x.
Apply Pytagora's equation to both triangles.
1) h^2 = 7^2 - x^2
2) h^2 = 8^2 - (6-x)^2
Eaual 1 to 2
7^2 - x^2 = 8^2 - 6^2 +12x -x^2
12x = 7^2 -8^2 +6^2 = 49 -64 + 36 = 21
x = 1.75
Substitue x-value in 1
h^2 = 49 - (1.75)^2 = 45.94
h = sqrt(45.94) = 6.78
Answer: option d.
Answer:
A, B, & D are polygons of equal straight lined sides. F is an irregular polygon which has unequal straight lined sides. C & E are NOT polygons as they have curves which cancels their polygon status.
