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DerKrebs [107]
3 years ago
7

Jk,kl, and lj are all tangent to circle o, ja=13,al=7, and ck=10 what is the perimeter of angle JKL

Mathematics
1 answer:
Alex73 [517]3 years ago
8 0

see the picture attached to better understand the problem


we know that

two tangent segments drawn from the same exterior point are congruent

so

JA=JB ,

LA=LC,

KC=KB

JA=13 units

LA=7 units

kC=10 units

hence

perimeter = JA+JB+LA+LC+KC+KB------> 13+13+7+7+10+10------> =60 units


therefore


the answer is

the perimeter of triangle JKL is 60 units

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Answer:

y=-\frac{1}{3}x-5

Step-by-step explanation:

Given:

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(x_1,y_1)=(-6,-3)\\(x_2,y_2)=(6,-7)

Now, for a line with two points on it, the slope of the line is given as:

m=\frac{y_2-y_1}{x_2-x_1}

For the points (x_1,y_1)=(-6,-3)\ and\ (x_2,y_2)=(6,-7), slope is:

m=\frac{-7-(-3)}{6-(-6)}\\m=\frac{-7+3}{6+6}\\m=\frac{-4}{12}=-\frac{1}{3}

Now, for a line with slope 'm' and a point (x_1,y_1) on it is given as:

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Plug in all the values and determine the equation of the line. This gives,

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Therefore, the equation of the line is:

y=-\frac{1}{3}x-5

4 0
3 years ago
Write an equation for the quadratic graphed below: x-intercepts: (-1,0) and (4,0); y-intercept: (0,1)
Luda [366]

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Step-by-step explanation:

The x-intercepts of the quadratic equation are simply it's roots.

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Now, formula for quadratic equation is;

y = ax² + bx + c

Where c is the y intercept.

At y-intercept: (0,1), we have;

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4a + b = -1/4 - - - (2)

From eq 1, b = -1 - a

Thus;

4a + (-1 - a) = -1/4

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3a - 1 = -1/4

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b = -5/4

Thus;

y = (1/4)x² - (5/4)x + 1

6 0
2 years ago
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5 0
3 years ago
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Answer:

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Step-by-step explanation:

8 0
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