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3 years ago

Line CD passes through points C(3, –5) and D(6, 0). What is the equation of line CD in standard form?

2 answers:

3 0

$\text{The formula of a slope:}\ m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{We have:}\\C(3;\ -5)\to x_1=3;\ y_1=-5\\D(6;\ 0)\to x_2=6;\ y_2=0\\\\\text{substitute}\\\\m=\dfrac{0-(-5)}{6-3}=\dfrac{5}{3}\\\\\text{The point-slope formula:}\ y-y_1=m(x-x_1)\\\\\text{substitute}\\\\y-(-5)=\dfrac{5}{3}(x-3)\\\\y+5=\dfrac{5}{3}x-5\ \ \ |-5\\\\\boxed{y=\dfrac{5}{3}x-10}$

NISA [10]3 years ago

3 0

= 5/2 x- 10.....................

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On the first visit to the veterinarian's office, Cora's kitten weighed 550 grams. On the second visit, the kitten had gained 350

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Answer:

On first visit, weight =550 g

Weight gained on second visit= 350 g

Weight gained on third visit= 300 g

Present weight=550+350+300= 1200

Weight gained=1200-550=650

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So the kitten gained 118% weigh

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Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths. X2 y2 − 2x 2y −

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Circles (represented by their equations in general form) in ascending order of their radius lengths can be arranged as first-fifth-second-forth-third-seventh-sixth.

<h3>What is the equation of circle?</h3>

The equation of the circle is the equation which is used to represent the circle in the algebraic equation form with the value of center point in the coordinate plane and measure of radius.

The standard form of the equation of the circle can be given as,

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

Here (h,k) is the center of the circle and (r) is the radius of the circle.

Lets arrange the first equation, in standard form of equation of circle as,

$x^2 +y^2 - 2x+ 2y -1 = 0 \\(x^2 - 2x) +(y^2+ 2y )= 1\\(x^2 - 2x+1) +(y^2+ 2y +1)= 1+1+1\\(x-1)^2+(y+1)^2=(\sqrt{3})^2$

The radius of this circle is √(3).

Lets arrange the second equation, in standard form of equation of circle as,

$x^2 +y^2 - 4x+ 4y - 10 = 0 \\(x^2 - 4x) +(y^2+ 4y )= 10\\(x^2 - 4x+4) +(y^2+ 4y +4)= 10+4+4\\(x-2)^2+(y+2)^2=(\sqrt{18})^2$

The radius of this circle is √(18).

Lets arrange the third equation, in standard form of equation of circle as,

$x^2 +y^2 - 8x - 6y - 20 = 0 \\(x^2 - 8x) +(y^2-6y )= 20\\(x^2 - 8x+16) +(y^2- 6y +9)= 10+16+9\\(x-4)^2+(y-3)^2=(\sqrt{45})^2$

The radius of this circle is √(45).

Similarly, radius of the forth, fifth, sixth and seventh circle is √(23), √(5), √(117) and √(46) units.

Hence, the circles (represented by their equations in general form) in ascending order of their radius lengths can be arranged as first-fifth-second-forth-third-seventh-sixth.

Learn more about the equation of circle here;

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