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Nonamiya [84]
3 years ago
9

How to write the fraction 6/6 as a sum of fractions three different ways

Mathematics
2 answers:
Aleks [24]3 years ago
6 0

3/6 + 3/6 = 6/6
2/6 + 4/6 = 6/6
1/6 + 5/6 = 6/6
kondaur [170]3 years ago
6 0

Answer:

The given fraction is

\frac{6}{6}

All fractions which sum 6/6 can be equivalent, as follows

<h3>1.</h3>

\frac{1}{6}+\frac{5}{6}=\frac{6}{6} is a way to represent the given fraction as a sum.

<h3>2. </h3>

\frac{4}{6} +\frac{2}{6} = \frac{6}{6}

<h3>3. </h3>

\frac{6}{12} +\frac{6}{12} =\frac{12}{12}=\frac{6}{6}

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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})

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