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:
0.30
Step-by-step explanation:
Probability of stopping at first signal = 0.36 ;
P(stop 1) = P(x) = 0.36
Probability of stopping at second signal = 0.54;
P(stop 2) = P(y) = 0.54
Probability of stopping at atleast one of the two signals:
P(x U y) = 0.6
Stopping at both signals :
P(xny) = p(x) + p(y) - p(xUy)
P(xny) = 0.36 + 0.54 - 0.6
P(xny) = 0.3
Stopping at x but not y
P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06
Stopping at y but not x
P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24
Probability of stopping at exactly 1 signal :
P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30
The answef is 1.83 because if you notice all of them are being divided by 38your answer is 16.5 because youre dividing all dog food in cups by wet food 3 times like 4.5/3 is 1.5 and so on
-7.08 as a fraction in simplest form would be -177/25
<h3>
Answer: (3, 0)</h3>
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Explanation:
Let's isolate x in the first equation.
x-2y = 3
x = 3+2y
Then we'll plug this into the second equation
Afterwards, solve for y.
2x + 4y = 6
2(3+2y)+4y = 6
6+4y+4y = 6
8y+6 = 6
8y = 6-6
8y = 0
y = 0/8
y = 0
Use this to find x.
x = 3+2y
x = 3+2(0)
x = 3
The solution is therefore (x,y) = (3, 0)
If you were to graph both lines, then they would intersect at the location (3,0).
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Checking the answer:
Plug x = 3 and y = 0 into the first equation.
x-2y = 3
3-2(0) = 3
3 - 0 = 3
3 = 3 that works
Repeat for the other equation
2x+4y = 6
2(3) + 4(0) = 6
6 + 0 = 6
6 = 6 that works as well
Both equations are true when (x,y) = (3,0).
The solution is confirmed.
