Answer:
The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.
Step-by-step explanation:
This is a optimization with restrictions problem.
The restriction is that the perimeter of the square cross section plus the length is equal to 108 inches (as we will maximize the volume, we wil use the maximum of length and cross section perimeter).
This restriction can be expressed as:
being x: the side of the square of the cross section and L: length of the package.
The volume, that we want to maximize, is:
If we express L in function of x using the restriction equation, we get:
We replace L in the volume formula and we get
To maximize the volume we derive and equal to 0
We can replace x to calculate L:
The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.
Answer:
Never
Never
Never
Step-by-step explanation:
The equations given are
2x1−6x2−4x3 = 6 ....... (1)
−x1+ax2+4x3 = −1 ........(2)
2x1−5x2−2x3 = 9 ..........(3)
the values of a for which the system of linear equations has no solutions
Let first add equation 1 and 2. Also equation 2 and 3. This will result to
X1 + (a X2 - 6X2) - 0 = 5
And
X1 + (aX2-5X2) + 2X3 = 8
Since X2 and X3 can't be cancelled out, we conclude that the value of a is never.
a unique solution,
Let first add equation 1 and 2. Also equation 2 and 3. This will result to
X1 + (a X2 - 6X2) - 0 = 5
And
X1 + (aX2-5X2) + 2X3 = 8
The value of a = never
infinitely many solutions.
Divide equation 1 by 2 we will get
X1 - 3X2 - 2X3 =3
Add the above equation with equation 3. This will result to
3X1 - 8X2 - 4X3 = 12
Everything ought to be the same. Since they're not.
Value of a = never.
Answer:
Step-by-step explanation:
so add what you have if you one number spilt it and divide
24 inches hope this helps
9514 1404 393
Answer:
see attached
Step-by-step explanation:
Polynomial long division is done the way any long division is done. Find a "partial quotient", subtract from the dividend the product of that partial quotient and the divisor. The result is a new dividend. Repeat until the degree of the dividend is less than that of the divisor.
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In the attached, the "Hints" show you how the partial quotient is found, and they show you how the product of the partial quotient and divisor is found.
The partial quotient term is simply the ratio of the highest degree terms of dividend and divisor. (Unlike numerical long division, there is no guessing.)
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The remainder is the dividend of lower degree than the divisor. As in numerical long division, the full quotient expresses the remainder over the divisor.
For example, 5 ÷ 3 = 1 r 2 = 1 + 2/3.
Your full quotient is (n+5) +1/(n-6).
