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2 years ago

9

Find the mean absolute value of 2, 5, 7, 10, 16 and 25, 29, 30, 31, 35 HURRRYYYY PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS

SSSSSSSSSSSSSSSS

1 answer:

telo118 [61]2 years ago

6 0

Answer:

mean = 19

i dont know what is meant by 'aboslute' mean so i just wrked out the normal mean

Step-by-step explanation:

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I need help on number 2 please help me. Thanks. Happy holidays

Airida [17]

Here, Pages in books = Number of books * number of pages in one book

So, Pages of picture book = 10 * 25 = 250
& Pages of chapter book = 120 * 5 = 600

So, Number of Total pages = 250 + 600 = 850

So, Option (D) is your answer! Hope this helps!!

5 0

3 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

Please help I need this for a text ASAP.

Sladkaya [172]

The values of x and y are 45 and 8.

<h3>What is a Linear Pair of Angles?</h3>

When two lines meet at a single point, a pair of linear angles is created. If the angles follow the point where the two lines cross, they are considered to be linear. A linear pair's angles add up to 180° in every case.

Given:

Measures of two linear pairs of angles which are of equal measures are 2x and 11y + 2.

As the sum of linear pair of angles is 180°, and if they are of equal measures, each angle must be equal to 90°.

To determine the values of x and y we equate the given angles to 90°.

2x = 90°

⇒ x = 45

And 11y+2 = 90°

⇒ 11y = 90-2 = 88

⇒ y = 8

To know more about linear pairs visit:

brainly.com/question/17525542

#SPJ13

8 0

1 year ago

What is the slope of a line that passes through the points. (-2, 3) and (4, -12)?

Crank

Answer:

5/2

Step-by-step explanation:

5/2

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3 years ago

A produce stand sells roasted peanuts for $1.90 per pound

goblinko [34]

Answer:

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Step-by-step explanation:

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