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natali 33 [55]
3 years ago
5

Simplify. x+5/x^2+6x+5

Mathematics
2 answers:
Alla [95]3 years ago
6 0

Answer:

The correct option is 2.

Step-by-step explanation:

The given rational expression is

\frac{x+5}{x^2+6x+5}

Factorize the denominator.

The middle term can be written as 5x+x.

\frac{x+5}{x^2+5x+x+5}

\frac{x+5}{x(x+5)+1(x+5)}

\frac{x+5}{(x+5)(x+1)}

Cancel out the common factors.

\frac{1}{x+1}

The simplified form of the given expression is \frac{1}{x+1}. The factors of the denominator are (x+5) and (x+1), it means at x=-5 and x=-1, the value of denominator is zero.

The given expression is not defined for x=-1 and x=-5.

Therefore the correct option is 2.

LuckyWell [14K]3 years ago
3 0
X+5/x^2+6x+5
x+5/(x+5)(x+1)
1/x+1
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Select interior, exterior, or on the circle (x - 5) 2 + (y + 3) 2 = 25 for the following point. (2, 3)
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Standard equation of a circle: <em>(x-h)² + (y-k)² = r²</em> where <em>(h, k)</em> is the center and <em>r </em>is the radius. In the case of our equation here, <em>(x-5)² + (y+3)² = 25</em>, we can conclude that our circle has a center at (5, -3) and a radius of 5 units.

We can use the distance formula with the center (5, -3) and our point (2, 3) to see how far away they are...if the distance between them is less than the radius of the circle, it is on the interior. If it's equal, it's on the circle. If it's greater, it's on the exterior.

Distance = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

Distance = \sqrt{(-3-3)^2+(5-2)^2}

Distance = \sqrt{(-6)^2+3^2}

Distance = \sqrt{36+9}

Distance = \sqrt{45}\approx6.7082

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If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle
stira [4]

Answer:

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<u>Step-by-step explanation:</u>

Given points end  Points are p(-3,-2) and q( 1,6)

<em>The distance of two points formula</em>

P Q = \sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2}   }

P Q = \sqrt{1 - (-3)^{2} + ((6 -(-2))^{2}   }

P Q = \sqrt{16+64} = \sqrt{80}

<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

<em>                        </em>(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )<em></em>

<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

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