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bekas [8.4K]
3 years ago
5

Find the midpoint of the line segment with endpoints (-9,-1,) and (7,0)

Mathematics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

The midpoint is;

(-1,-0.5)

Step-by-step explanation:

To do this, we shall use the midpoint formula

That would be;

(x,y) = (x1 + x2)/2 , (y1 + y2)/2

(x1,y1) = (-9,-1)

(x2,y2) = (7,0)

(x,y)

= (7-9)/2 , (0-1)/2

= -2/2 , -1/2

= (-1, -0.5)

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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
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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:

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Make x the subject

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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

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

Different A with respect to y

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

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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

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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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