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3 years ago

Are there more or less than 5 pt of blood in an adult human?

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

5 0

There are more than 5 pints of blood in an adult human. the average is approximately 10 pints

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F(x) = 3x -6 What is 7 times f(x)

Alina [70]

Answer:

21x - 42, see below :)

Step-by-step explanation:

Hello!

We know that f(x) is equal to 3x - 6

We want to know what 7 * f(x) is so we should first substitute f(x) for 3x - 6

7 * f(x)

7(3x-6)

We use distributive property to open the parenthesis:

7 * 3x + 7 * -6

21x - 42

8 0

3 years ago

Read 2 more answers

Come up with an example for a solid, liquid, and gas describing the particle movement.

Volgvan

Answer:

Ice Cube, Liquid Water, and Steam

Step-by-step explanation:

An example of a solid is a ice cube, the particles are compacted together and there's very little movement.

An example of a liquid is water, the particles are more spread around and the movement is more frequent.

An example of a gas is steam, the particles are anywhere, moving fast and sporadic.

3 0

2 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

Fill in the blanks below.

Dmitry_Shevchenko [17]

Answer:

a) Slope of the line m = ∞

b) Slope of the line m =0

Step-by-step explanation:

Explanation:-

Step(i):-

Given points are ( 7,7) and (7,-8)

Slope of the line

$m = \frac{y_{2} - y_{1} }{x_{2}-x_{1} }$

$m = \frac{-8-7 }{7-7 } = \frac{-15}{0} = infinite$

m = ∞

Step(ii):-

Given points are (-6,7) and (9,7)

Slope of the line

$m = \frac{y_{2} - y_{1} }{x_{2}-x_{1} }$

$m = \frac{7-7 }{9-(-6) } = \frac{0}{15} = 0$

Slope of the line m =0

6 0

3 years ago

(−2)⋅(8)−(4)⋅(−3) A. -17 B. -3 C. -4

Gwar [14]

The Answer To This Problem is: C.-4

8 0

3 years ago