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3 years ago

14

The question is in the picture and the answer choices are A.) (t,0)B.) (-r,0)

1 answer:

VladimirAG [237]3 years ago

8 0

the answer is a hope this help

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3 years ago

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Step-by-step explanation:

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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:

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

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2 years ago

A: x/4 +1 = -3 B: x+4 = -12 1) How can we get Equation B from Equation A?

Sveta_85 [38]

Answer:In the equation y = mx + b for a straight line, the number m is called the slope of the line. Let x = 0, then y = m • 0 + b, so y = b. The number b is the coordinate on the y-axis where the graph crosses the y-axi

Step-by-step explanation:

5 0

3 years ago

Please help me answer this I don't understand, I would really appreciate how you solved it as well.

abruzzese [7]

Answer: Parallel

Step-by-step explanation:

Set your -2x+10y=5 equation equal to y

-2x+10y=5

<u> -10y -10y</u>

-2x=5-10y

<u>-5 -5 </u>

<u>-2x-5</u>+<u>-10y</u>

-10 -10 -10

-1/5x-1/2=y

Now you can put your equations into your graphing calculator and examine the lines made

(if you dont have one search up <em>online graphing calculator</em> on google or your app store)

5 0

3 years ago