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3 years ago

15

What is square root of 674

2 answers:

laiz [17]3 years ago

8 0

Answer:

25.9615099715

Step-by-step explanation:

yulyashka [42]3 years ago

7 0

25.96 that’s the square root of your answer

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stiv31 [10]

You can solve mean by adding up all the numbers then dividing it by the number of numbers there are.

You can solve the median by putting all of the numbers in order and then crossing off one in the beginning then crossing off one at the end. You can continue that until you get to one number in the middle.

You can solve the mode by looking for the most frequent number. That's the mode. You can remember that by looking at the MO and remember most often.

I hope this helps :-)

6 0

4 years ago

How to solve for the shaded region

k0ka [10]

The shaded region is a triangle. Triangle area can be determined with formula
a = 1/2 × b × h

From the question, we know that the triangle consists of
base = 10 - 6 = 4 m
and the height is the dimension which is perpendicular to the base,
height = 6 m

Find the triangle area
a = 1/2 × b × h
a = 1/2 × 4 × 6
a = 1/2 × 24
a = 12

The area of the shaded region is 12 m²

5 0

3 years ago

9p- 18=27 what is p?

kozerog [31]

Answer:

p = 5

Step-by-step explanation:

4 0

3 years ago

PLEASE HELP ASAP<br><br>Solve. <br><br>y > 2x + 2<br>y ≤ -3/2 x - 5

Fynjy0 [20]

$2x + 2 < - \frac{3}{2} x - 5 \\ 2x + \frac{3}{2} x < - 5 - 2 \\ 3.5x < - 7 \\ x< - 2$
$x < - 2 \\ x < \frac{y - 2}{2}$
$-2 = \frac{y - 2}{2} \\ y=-2$

7 0

3 years 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>

3 0

3 years ago