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DerKrebs [107]
3 years ago
12

Geometry help. Inscribing circle help and constructing lines.

Mathematics
1 answer:
prohojiy [21]3 years ago
3 0
<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>

To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.

1. Draw the angle bisector of angle JKL. The straight line in the green one.

2. Draw the angle bisector of angle KLJ. The straight line in the blue one.

3. Draw point D at these two lines.

4. Draw a line through point D perpendicular to side JK. The straight line in the red one.

5. Mark point G at the red line and side JK.

6. Draw a circle with center at D (point of concurrency) and a radius of DG.


<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>

To circumscribe a circle in the attached triangle, we must follow the following steps:

1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.

2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.

3. The center of the circle is the intersection of these two lines

4. Draw a circle with center at D (point of concurrency).

<h3>3. Construct the two lines tangents to circle</h3>

A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.

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Which statement best reflects the solution(s) of the equation?
Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1} and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0

Now, factoring the term:

x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}  the denominator becomes zero i.e

\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2  in the equation,

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}

\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

  • brainly.com/question/1626495
  • brainly.com/question/2959656
  • brainly.com/question/2456302

#learnwithBrainly

5 0
4 years ago
Express sin A, cos A, and tan A as ratios ​
vlabodo [156]
<h3>Answer: Choice B</h3><h3>sqrt(3)/2,    1/2,   sqrt(3)</h3>

================================================

Explanation:

Sine of an angle is the ratio of the opposite side over the hypotenuse. For reference angle A, the opposite side is BC = 6sqrt(3). The hypotenuse is the longest side AB = 12

Sin(angle) = opposite/hypotenuse

sin(A) = BC/AB

sin(A) = 6sqrt(3)/12

sin(A) = sqrt(3)/2

---------------

Cosine is the ratio of the adjacent and hypotenuse

cos(angle) = adjacent/hypotenuse

cos(A) = AC/AB

cos(A) = 6/12

cos(A) = 1/2

---------------

Tangent is the ratio of the opposite and adjacent

tan(angle) = opposite/adjacent

tan(A) = BC/AC

tan(A) = 6sqrt(3)/6

tan(A) = sqrt(3)

5 0
3 years ago
Tell me which lines are parallel to each other.<br> -6x + 2y = 8; y = 4; y = 3x; y = 3
Sloan [31]

Answer: -6x+2y=8 and y=3x

Step-by-step explanation:

they must have the same slope to be parallel and when you convert -6x+2y=8 into y=mx+b it is y=3x+4

6 0
3 years ago
A rectangle has an area of 12 feet and a length of 5 feet .what is its width
Reil [10]
Area= l * b

12= 5 * b

b = 12/ 5

b= 2.4 ft
8 0
4 years ago
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viva [34]
The 3 in 350 is in the hundred place and in 403 the 3 is in the ones place
3 0
3 years ago
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