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
Well, there isn’t really an end for numbers...
However; The biggest number referred to regularly is a googolplex (10googol), which works out as 1010^100. That isn’t the end to numbers but it is a huge one. We will replace that with ‘all the numbers in the world’.
106 is the exponent equivalent to 1 million
So your question would be:
106 x 1010^100 =
However I don’t believe there is a calculator that large.
Answer:
x=4
Step-by-step explanation:
We can find the length of RT by using the Pythagorean theorem
sin theta = opposite side/ hypotenuse
sin 60 = 2 sqrt(3)/ RT
Multiply each side by RT
RT sin 60 = 2 sqrt(3)
Divide by sin 60
RT = 2 sqrt(3)/ sin 60
RT = 4
Then
tan theta = opposite side/ adjacent side
tan 45 = x/RT
tan 45 = x/4
4 tan 45 = x
4 = x
