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1 year ago

9

Cynthia is standing at the center of a circular playground. Which statement is true with respect to Cynthia’s position?

2 answers:

sergeinik [125]1 year ago

7 0

D. The distance from Cynthia to pole 3 us equal to the distance from Cynthia to pole 1.
pole one and three are like the same distance apart from Cynthia.

garri49 [273]1 year ago

7 0

D. The distance from Cynthia to pole 3 us equal to the distance from Cynthia to pole 1.

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Lila needs a custom frame for a painting that her friend made for her. I’ve Been Framed is offering 50 percent off plus an addit

nikklg [1K]

Answer:

The location offering the better price is The Framers. It is cheaper by $1.35

Step-by-step explanation:

I've Been Framed:

50% of $115 = $57.5

10% of $57.5 = $5.75

$57.5 - $5.75 = $51.75

Total: $51.75

The Framers:

30% of $120 = $36

$120 - $36 = $84

40% of $84 = $33.6

$84 - $33.6 = $50.4

Difference: $1.35

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1 year ago

Need help RN!!!!

Vesnalui [34]

Answer:

log y = 3

Step-by-step explanation:

Substitute 10 for x, obtaining the following:

log y = 0.23(10) + 0.8, or

log y = 2.3 + 0.8, or 3.10, or (to the nearest whole number):

log y = 3

Step-by-step explanation:

Hope this helps:)

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1 year ago

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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The answer is fifty percent.

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