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2 years ago

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LK is tangent to circle J at point K. Circle J is shown. Line segment J K is a radius. Line segment K L is a tangent that inters

ects at point K. A line is drawn from point L through a point on the circle to the center point J. The length of the radius is r, the length of K L is 11, and the length of the line segment from point L to the point on the circle is 6. What is the length of the radius? StartFraction 6 Over 85 EndFraction StartFraction 85 Over 12 EndFraction StartFraction 121 Over 36 EndFraction StartFraction 157 Over 12 EndFraction

2 answers:

Evgen [1.6K]2 years ago

8 0

Answer:

$(B)r=\dfrac{85}{12}$

Step-by-step explanation:

Given a circle centre J

Let the radius of the circle =r

LK is tangent to circle J at point K

From the diagram attached

LX=6
Radius, XJ=JK=r
LK=11

Theorem: The angle between a tangent and a radius is 90 degrees.

By the theorem above, Triangle JLK forms a right triangle with LJ as the hypotenuse.

Using Pythagoras Theorem:

$(6+r)^2=r^2+11^2\\(6+r)(6+r)=r^2+121\\36+6r+6r+r^2=r^2+121\\12r=121-36\\12r=85\\r=\dfrac{85}{12}$

The length of the radius, $r=\dfrac{85}{12}$

nadezda [96]2 years ago

5 0

Answer:

The answer is B on Edge 2020

Step-by-step explanation:

I did the assignment

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F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

34kurt

Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

$f(x) = 3x^2 + 9$

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

<em>Hence, f(x) and g(x) and not inverse functions</em>

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