A. You can remember which axis is which by Y-axis is the vertical one (Y to the sky) and X- axis is horizontal. it's not C because you don't have a second point. It's not (0,4) it's just 4.
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
Answer:
60%
Step-by-step explanation:
Answer:
300 i think hope it helps
Answer:
y = 45 / 2x + 50
Step-by-step explanation:
The best equation for the line of the best fit would be y = 45 / 2x + 50.
The slope would be 45 / 2 and the y-intercept would be ( 0 , 50 ).
I used the points, ( 0 , 50 ) and ( 4 , 140 ) to get the equation, slope, and the y-intercept.
I used those points because they were the most accurate ones that I know for now.
Also, remember the formula is, "y = mx + b."
