Down, the line is going down
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:
See explanation
Step-by-step explanation:
cos(θ)= 18/30 = 3/5
sin(θ)= 24/30 = 4/5
tan(θ)= 24/18 = 4/3
cos(ϕ)= 24/30 = 4/5
sin(ϕ)= 18/30 = 3/5
tan(ϕ)= 18/24 = 3/4
You could do the question the way it is written, but it is far easier to bring the negative power up to the numerator.
y= x^2 - 3. The derivative of that is
dy/dx = 2x The three is a constant and is always dropped when a derivative is taken
d(-3)/dx = 0
If you are a purist and want to solve the question the way it is written, you could do it this way.
dy/dx = d(1)/dx x^-2 - d(x^-2)/dx * 1
======================
(x^-2)^2
dy/dx = - (-2 x^ - 3) / x^-4
dy/dx = 2 x^-3 * x^4
dy/dx = 2 x^(-3 + 4)
dy/dx = 2x ^ 1
dy/dx = 2x <<<<< answer
Answer:
soln
Step-by-step explanation:
a=7
b=-2
5a-3b
5×7-3×-2
35+6(because - and - becomes+
41ans
