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Finger [1]
2 years ago
14

Please help with 24-26

Mathematics
2 answers:
MrMuchimi2 years ago
4 0

20. - 23. are all correct.

24. A number raised to the zero power equals 1. 0 to the 0 power may be undefined according to some, but any number other than zero raised to the zero power is 1. We assume x is not equal to zero.

-2x^0 = -2 \times 1 = -2

25. To raise a product to a power, raise each factor to the power. Assume that none of the variables equal zero.

(11abx)^0 = 11^0 \times a^0b^0x^0 = 1 \times 1 \times 1 \times 1 = 1

26. Assume all variables are non-zero.

(xy)^0 - (pq)^0 = 1 - 1 = 0

Inga [223]2 years ago
3 0

24

The power 0 only affects the x. x^0 = 1

-2 * 1 = -2

25

Here everything inside the brackets is affected by the 0.

(11abx)^0 = 1 There is nothing left over.

26

(xy)^0 = 1

(pq)^0 = 1

1 - 1 = 0


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a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

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Step-by-step explanation:

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(x, y): (\frac {1} {2}, \frac {9} {2})

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