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3 years ago

9(x) = 2x-1 r(x) = -2x 2 2 Find the value of r(q (2)).

Mathematics

1 answer:

BabaBlast [244]3 years ago

6 0

Answer:

Answer is -6

Step-by-step explanation:

$given \: q(x) = 2x - 1 \\ and \: r(x) = - 2x \\ r(q(2)) = r(2(2) - 1) \\ \: \: = r(4 - 1) \\ = r(3) \\ = - 2(3) \\ = - 6 \\ therefore \: r(q(2)) = - 6$

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Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

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Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

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$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

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

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

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The equation of the circle is given by:

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

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

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We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

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