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3 years ago

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1. A cube-shaped aquarium has edges that are 4 ft. long. The aquarium is filled with water that has a density of

1 answer:

mr_godi [17]3 years ago

3 0

So first you find the volume of the cube which is 4x4x4 or 64
(a) is asking what the cube ways with water so you multiply 51 and 64= 3264 so the table would break so, no
(b) the equation 51 lb per cubic feet is the density so would filling it half way change the equation? also no

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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3 years ago

Which statement is true about the equations 3x + 4y = 12 and ¿-X- = = 1?

jarptica [38.1K]

Answer:

cfhjytgfhe5tgnbg5etjgnbg345jthgbyu6jtngfyu6jyh

Step-by-step explanation:

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3 years ago

Help pls is geometry

Burka [1]

Answer:

C) 16, 6

Step-by-step explanation:

Set AB and DC equal to eachother. 4x = x + 12.
Subtract x from both sides. 3x = 12
Divide by 3 to get x alone. x = 4
Plug this x value in the equation for AB. 4•(4) = 16
We know the AD equals 6, so that will be one of the values and we now know that AB equals 16.

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3 years ago

The height of the pyramid in the diagram is three times the radius of the cone. The base area of the pyramid is the same as the

Brut [27]

Answer:

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Step-by-step explanation:

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3 years ago

How to calculate 99 : 5

Goryan [66]

Answer:

19.8

Step-by-step explanation:

99/5= 19.8

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3 years ago