Construct a line that is tangent to a circle at a given point

1 answer:

6 0

Steps:

1) Draw the circle and mark the center (A) and the given point (B)

2) Draw a line that passes through the center (A) and the given poiint (B).

3) Draw a perpendicular line to the segment AB.

The last line drawn is tangent to the circle at the given point.

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What is the difference? startfraction x over x squared minus 2 x minus 15 endfraction minus startfraction 4 over x squared 2 x m

prisoha [69]

The difference is StartFraction x squared + 3 x minus 12 Over (x + 3) (x minus 5) (x + 7) EndFraction.

<h3>What is a quadratic equation?</h3>

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

The given expression is

$\frac{x}{x^{2} -2x-15} -\frac{4}{x^{2} +2x-35}$

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by Taking LCM

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$\frac{x^{2} +3x -12}{(x-5)(x+3)(x+7)} \\$

Learn more about a quadratic equation;

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2 years ago

Graph y+6=4/5(x+3) using the point and slope given in the equation

Elodia [21]

Slope is 4/5
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Your first point should be (3,6) then from that point go up 4 and 5 to the right, that should be your second point, then graph.

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Solving quadratics using square root:<br>12 + (x-2)^2= 3

jeka94

$\bf 12+(x-2)^2=3\implies (x-2)^2=-9\implies \sqrt{(x-2)^2}=\sqrt{-9}
\\\\\\
x-2=\sqrt{-9}\implies x=\sqrt{-9}+2$