1. x = 25 inches
2. Triangle c
1 figure = $17 so you multiply 4 and 17 to get 68
Martin will need to save $68
Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
There are two ways to list the angles:
1) Simply name them based on the points:
∠W , ∠X , ∠Y , ∠Z
2) The way that I believe that you are supposed to list them in this case. List them as such:
∠WYZ , ∠YZX , ∠ZXW , ∠XWY
~
Answer: It is TRUE. A point in a linear equation can be the solution to said equation.
