Student 1 Mrs. Nomial is standing about 65.2 feet from the edge of the carousel.
Student 2
Student 3 the mother is standing
130 ft from the center the mother is standing
130 ft from the center
1)From the graph, i can tell that Mrs. Nomial is standing at the point (130, 18).
2)The diameter of the carousel is 130 feet, the radius is 65 feet. I graphed the carousel and Mrs. Nomials points on graph paper.
3)I found the line that Mrs. Nomial is standing on, i found the slope which was 20/130 or .1538. The line was Y=.1538X.
4)This line ran through the point which is closest to her. It intersected the circle at (64,8).
5)I subtracted (66,8) from (130,20). I then used the distance formula to find the distance between the points. After solving
D= sqrt(64^2+12^2)
D= sqrt(4240)
D= 65.2
6)I now know that Mrs. Nomial is standing about 65.2 feet from the carousel. i wasnt able to understand what to do so i asked my father and from what he understood (from me) is how he showed me to figure out hte answer and i figured that it was 130 feet because the whole ride is
130 feet across and half of the would be 65 feet... and from the center where the "first" 65 start i counted on to where the mother was and for each unit i added 10 feet which came to 130 feet...where the children's mother was standing watching them - irini i wasnt able to understand what to do so i asked my father and from what he understood (from me) is how he showed me to figure out hte answer and i figured that it was 130 feet because the whole ride is
130 feet across and half of the would be 65 feet... and from the
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Student 4
Student 5
Student 6
Student 7
Student 8
Student 9 edfgadsgsfdhdfgdasf
Mrs. Nomial is standing 66 feet away from the carousel.
Ms Nomial was about
275 feet away from the carousal. center where the "first" 65 start i counted on to where the mother was and for each unit i added 10 feet which came to 130 feet...where the children's mother was standing watching them - irini dfadsfdsfdsfadsfdasfdsfadsfasdfadsfada fdag fdagdag adfgafdgfd dafgafgvafgadfg fdg adfg
First I found the slope, then I found the radius, next I ofund the equatio and the equation for the two crossing lines. After all that i found the distance from Mrs. Nomial to the carousel.
Using properties of sine cosine and tangent I found the angles of a triangle with the tangent line, radius and a hypotenuse from the center to the point where ms nomial was standing which met with the tangent line. I now had a 90 degree triangle. To find the first angle of the triangle i had to go back a step and make an angle between the two radii. Making another right triangle, i found that it was 22, subtracted it from 180 giving me 158 degrees. Since the hypotenuse line from the center of the circle bisected the angle, to find my first angle of the triangle i divided the angle by 2 which gave me 79 degrees. and subtracting 79 and 90 from 180, i found the third angle, which was 11 degrees. And of course, the radius side was already known, 65. Using law of tangent, i found the hypotenuse which was 340 and through the pythagorean theorem i found the third side and checked my answer. The hypotenuse gave me the distance from
Ms Nomial to the center of the carousal, i needed her distance from the closest part of it. So i simply subtracted 65 (the radius length) from her distance to get her distance from the edge of the carousal.
Since that each point represented 10 feet and i went on and counted up to where Mrs. Nominal would be at on the x-axis which was to the
45 point and just estimated where she'd be.
She would be 50 feet away, for the extra i'd be impossible for mrs. nominal to see them both.
I think Nomaial is about 66 feet from the carousel. my final answer was
42.18
First i found the slopes of the radius segments. Then i found the line of intersection of the two lines of site. Then i found the distance from Mrs. Nomial to the carousel.
First I found the equations of Mrs. Nomial's lines of sight using the negative reciprocal of the slopes of both radiuses to the children. Using those line equations, I found the points of intersection to find where Mrs. Nomial was standing. Then I calculated her distance from the center (0,0) and subtracted from it the distance from the center to the outer edge (65). That number was
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Student 10
Student 11
Student 12
She is 58.5235 ft. from the closest point of the carousel. my final answer
I forgot to add the picture.
I guess I read the question wrong the first time, but now, when I needed to calculate how far Mrs. Nomial was from the carousel,
I first made a triangle from where she was to the closest point of the carousel. (Look at the attachment.) Then what I did was find out the distances, plug them into the Pythagorean Theorem since my triangle was a right triangle. Then when I added 55 squared and 20 squared, I got 3425. Then I found the square root of that and got
58.5235 feet. distance formula Mrs. Nomial is standing 66.53 feet from the carousel
Mrs. Nomial is standing 129.7 feet from the center of the carousel.
I noticed that the points of Mrs. Nomial to her daughter at 25, 60 to the center of the carousel creates a right triangle because the definition of a perpendicular line says that it is 90 degrees. I can then find the length of the sides of this triangle. I know that this distance from the center of the carousel to the first daughter at 25,60 is 65 feet. This is because the diameter of the carousel is 130 feet, and so the distance to the both daughters from the center is 65 feet. The daughters have two congruent triangles that share the side formed from the center of the carousel to their mother. Also, since they are tangent segments drawn from the same point (Mrs. Nomial), the distance from each daughter to Mrs. Nomial is the same.
So we have the length of one side. To find the length of the side connecting Mrs. Nomial and her daughter, I used the distance formula. I looked at the grid and found that Mrs. Nomial's coordinates are about 130,20 and it was given that one of her
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Student 13
Student 14
The length of h is
131.52 so the the diamiter is 130 right.
Half of the diamiter is the radius= 65. using distance formula I got
66.52 as the distance from the edge of the caracell to mis nomial.
Mrs. Nomial is about
66 feet from the edge of the carousel. daughter's coordinates is 25,60. So I used these numbers in the distance formula:
d= sqrt(x2-x1)^2 + (y2-y1)^2
= sprt(130-25)^2 + (20-60)^2
=sqrt(11025) + (1600)
=sqrt(12625)
=112.36 ft
Now I have the lengths of two sides of two congruent right triangles so I can use the pythagorean theorem to find the missing side that is common to both.
a^2 + b^2 = c^2
(65)^2 + 112.36)^2 = c^2
4225 + 12624.7696 = c^2
16849.7696 = c^2
sqrt(16849.7696) = c
129.8 ft= c
This shows that the distance from Mrs. Nomial to the center of the carousel is 129.8 ft. You could also say that she is 64.8 ft away from the edge of the carousel, but I don't think that is important.
I'm pretty sure that my answer is correct, although I think I could have used a simpler strategy. I also could have just used the distance formula to find the distance between Mrs. Nomial and the center by using the 0,0 coordinates. However, I was hoping that I would get a better idea of how to solve this while going the long way. so using distance fourmula its the distance from the center of the caracell to where mis nomial is. so d= sqart(130-0)^2+(0-20)^2 equals (131.52) the distance from mis nomial to the center of the caracell. the distance from the edge to mis nomial is 66.52 (thats the raidius - the answer you got for the problem you solved for the distance formula.so 65-131=66.52.
I saw there were two lines in the problem - one from Mrs. Nomial to each daughter. I got an equation for each line and then figured out where the two lines intersected to see where Mrs. Nomail is standing.
Then I used the distance formula to figure out how far away she is.
To get the equation of the lines, I found the slope of each radius
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© 1994-2020 Drexel University http://mathforum.org/pows/ going to the girls. Since one end of each radius is (0,0), the slope is just the y-value of the point where the girl is over the x-value.
So the top radius has a slope of 60/25 which reduces to 12/5. The bottom radius has a slope of -52/39 which reduces by 13 to -4/3.
It said in the problem that the lines are perpendicular to the radiuses, so the slopes of the lines are the negative reciprocals of the slopes of the radiuses. The top line has a slope of -5/12 and the bottom one has a slope of 3/4.
Now I know the slope and a point on the line for each line, so I used the point-slope form to find the equations. y - 60 -5 y - (-52) 3
------ = --- --------- = --- x - 25 12 x - 39 4
12(y - 60) = -5(x - 25) 4(y + 52) = 3(x - 39)
12y - 720 = -5x + 125 4y + 208 = 3x - 117
12y + 5x = 845 4y - 3x = -325
This is a system of two equations and two variables. I multiplied the second equation by -3 and then added the two equations to cancel out the y.
-12y + 9x = 975
12y + 5x = 845
---------------
14x = 1820
x = 130
Then I put x = 130 into the second equation to get y.
4y - 3(130) = -325
4y - 390 = -325
4y = 65
y = 16.25
Mrs. Nomial is standing at the coordinates (130, 16.25). That looked right on the picture, so I figured I was on the right track. Now I
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Student 15
Student 16
She was 66 ft away from the carousel
Mrs. Normial is 66 feet away from the carisel. uh oh have to find the distance, so I used the distance formula to find how far (130, 16.25) is from (0,0), the center of the carousel. d = SQR[(130 - 0)^2 + (16.25 - 0)^2]
= SQR[16900 + 264.0625]
= SQR[17164.0625]
= 131.01
So she's standing about 131 feet from the center of the carousel. The carousel is 130 feet across, so its radius is 65 feet. The line from the center to Mrs. Nomial represents a radius and then her distance from the edge of the carousel. I subtracted the radius from the 131 feet and got 131 - 65 = 66.
Mrs. Nomial is standing about 66 feet from the edge of the carousel. find the slopes of the radius segments and then use those slopes to determine the slopes of the lines of sight find an equation for each of Mrs. Nomial's lines of sight find the point of intersection of the two lines of sight find the distance from Mrs. Nomial to the carousel
First I looked up the description of a tangent line to make sure I know what it is. Then knowing that, I used an equation that consists of several prosecess. The first one is that x=x0+(a)cos(t) and y=y0+(A)sin(t). This is the center of the circle points that you plug in. So now you plug all the values in and next you multiply {x=x0+(a)cos(t) y=y0+(A)sin(t)} times {a cos t and a sin t) pluging in the varibles used in the problem in the propor place like x and y equal to 0. Then you use the quadraric formula and do it like this t=(+)(-) cos {-x (+)(-)y and square root (x^2+y^2-a^2)/
(x^2+y^2) and when you are done you get 66 feet. i dont know Student 17
Student 18 Mrs. Nomial is standing about 66 feet far from the carousel.
First, you draw a circle origin of the carousel with the point outside of the circle. Since you know that diameter of the carousel is 130, you divide 130 by 2 to get radius which is 65. Because the problem gives you that the tangent line is perpendicular to the radius at the point of the tangency, the angle they create is 90
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Student 19 my solution is 2 degrees. Next, you use the distance formula to find the distance between the given circle to the other circle you drew earlier. So the distance formula would be like d= square root of (39+130)^2-(-
52+15)^2. That formula gives you the distance of Mrs. Nomial stands from the carousel which is about 66 feet. i solved it and got 5
Student 20 the answer is 10 5 + 5 = 10
Student 21
Student 22
Student 23
Student 24
The kids are each 100 feet away from their mother.
Mrs. Nomial was standing about 139.3ft away from the edge of the carousel.
Nomial is about 66 feet from the carousel
Mrs. Nomial would be standing about 66.012 feet away in order to see both her daughters.
I measured it and we used are mad graphing skills.
NOTE: v/ = square root
First, I found the coordinates where the middle of the edge of the circle was and the coordinates were (65,0). Then I found the coordinates where Mrs. Nomial was standing and it was (15,130). Then
I used the distance formula to find how far away she was standing.
Here's the steps: d=v/(x2-x1)^2+(y2-y1)^2 d=v/(15-65)^2+(130-0)^2 d=v/(-50)^2+(130)^2 d=v/2500+16900 d=v/19400 d=139.3 ft
First I found the slopes of the radius segments and then the slopes to find the lines of sight
I found a equation for every Ms nomial;s lines of sight.
I tried to find the point of intersection of the two lines of sight
I dound the distance from ms nomial to the carousel
This problem was initially hard, but once I finally got going it was pretty easy. The first thing that I noticed was that a kite shape was formed by connecting the point where the mother was with the 2 points of tangency using the tangent lines to connect the points.
Then you would connect the points of tangency with the origin of the circle, between these points would be the radius of the circle. This shape could also be seen as 2 right triangles connected by their hypotenuses, with the hypotenuse running from the mother to the origin of the circle.
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Student 25 Mrs. Nomial was standing 68.9 feet from the carousel's outside border.
Student 26 The answer to the question is that Mrs.
Nomial is standing
22.59 feet away from the carousel.
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At first I did not know how this would get me to my answer, so I tried to find as much as I could about the shape.
The rectangular coordinates of the points of tangency were given in the problem, and I found that if I converted the points to polar coordinates I could get the length of the radius as well as the angle it is at. I converted (25,60) and (39,-52) using a graphing calculator and came up with polar coordinates of (65,
67.38°) and (65,-53.13°) respectively. This means that the radius is
65 feet long and the total measurement of the angle that is formed at the origin of the circle is 120.51°. The hypotenuse of the 2 triangles would bisect this measurement making 2 angles of 60.255°.
Since I then knew the measurement of 2 angles within the right triangle I could subtract their measurements from 180 and get the 3rd angle, the angle opposite the radius of 65 ft. 180 – 90 –
60.255 = 29.745°. This shows that the angle would be 29.745°.
Now that I had the angle opposite the radius, I could use the law of sines to determine the length of the hypotenuse. 65 / sin(29.745°) = c / sin(90°). Using this equation I cross multiplied, with the aid of a calculator, to determine that c = 131.012. Since this measurement is from the mother to the origin of the circle and not from the mother to the edge of the circle, I had to subtract the radius of 65 from 131.012 and got a final answer of 66.012.
Extra Credit: If the girls were at different ends of the circle, the tangent lines would be parallel and would never meat, meaning the mother would not be able to see both of them at the same time at any distance.
Mrs. Nomial is at a coordinate of (130,20). The tangent meets the circle at (25,60). The length of this tangent = the square root of
(25 - 130)squared + (60 - 20) squared. This is equal to 117.04 feet.
Then, using pythagoras theorum of right triangles the distance between Mrs. Nomial and the center of the carousel = the square root of 117.04 feet squared + 65 feet squared. This is equal to 133.9 feet, if you subtract the 65 foot radius of the circle you come up with 68.9 feet from the carousel.
The way I found my answer was by using the tangent that is created when the point are connected. First of all I had to figure out what the angles were. Since I know that a tangent line creates a perpendicular angle with the radius of the circle at the point of tangency. I also knew that a quadrilateral equals 360 degrees. So I
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Student 27
Student 28
Student 29
Student 30
Student 31
Student 32
Mrs.Nomial is standing
67 feet from the carousel. had 180 degrees I subtract that from 360 degrees and got 180 degrees.
I then subtracted 45 degrees because the point that Mrs.Nomial is standing creates a 45 degree angle. I then got 135. I divided that by
2 because the figure is a quadrilateral and in order to get a triangle the number needed to be divided by 2. When I did that I got
67.5 degree. I then used the calculator to figure out what the tangent of 67.5 is. The calculator screen had typed on it was tangent
(67.5),then I pressed enter and got 22.58818053. I then rounded it to
22.59. That is my answer 22.59 feet. one point is (25,60) and the other one is (39,-52). I tried to find the slopes of the coordinates given.
(39-25)^2+(-52-60)^2
196+12544
=112.87 since it was a triangle, and a triangle equals to 180, so I subtracted 180 to 112.87.
180-112.87
=67
Well i just used a2= c2+b2 She will ne 6 feet awayu
Mrs. Nomail is standing 65.365 feet away from the carousel
Short Answer d
I found the distance between the center and one girl, then the distance between that girl and the mom. I then plugged this into the pythagorium theorem daf 3
Long Answer
Mrs. Nomial is standing about 65.2 feet from the edge of the carousel.
1)From the graph, i can tell that Mrs. Nomial is standing at the point (130, 18).
2)The diameter of the carousel is 130 feet, the radius is 65 feet. I graphed the carousel and Mrs. Nomials points on graph paper.
3)I found the line that Mrs. Nomial is standing on, i found the slope which was 20/130 or .1538. The line was Y=.1538X.
4)This line ran through the point which is closest to her. It intersected the circle at (64,8).
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Student 33
Student 34
Student 35
Student 36
Student 37 the mother is standing
130 ft from the center the mother is standing
130 ft from the center edfgadsgsfdhdfgdasf
Mrs. Nomial is standing 66 feet away from the carousel.
Ms Nomial was about
275 feet away from the carousal.
5)I subtracted (66,8) from (130,20). I then used the distance formula to find the distance between the points. After solving
D= sqrt(64^2+12^2)
D= sqrt(4240)
D= 65.2
6)I now know that Mrs. Nomial is standing about 65.2 feet from the carousel. i wasnt able to understand what to do so i asked my father and from what he understood (from me) is how he showed me to figure out hte answer and i figured that it was 130 feet because the whole ride is
130 feet across and half of the would be 65 feet... and from the center where the "first" 65 start i counted on to where the mother was and for each unit i added 10 feet which came to 130 feet...where the children's mother was standing watching them - irini i wasnt able to understand what to do so i asked my father and from what he understood (from me) is how he showed me to figure out hte answer and i figured that it was 130 feet because the whole ride is
130 feet across and half of the would be 65 feet... and from the center where the "first" 65 start i counted on to where the mother was and for each unit i added 10 feet which came to 130 feet...where the children's mother was standing watching them - irini dfadsfdsfdsfadsfdasfdsfadsfasdfadsfada fdag fdagdag adfgafdgfd dafgafgvafgadfg fdg adfg
First I found the slope, then I found the radius, next I ofund the equatio and the equation for the two crossing lines. After all that i found the distance from Mrs. Nomial to the carousel.
Using properties of sine cosine and tangent I found the angles of a triangle with the tangent line, radius and a hypotenuse from the center to the point where ms nomial was standing which met with the tangent line. I now had a 90 degree triangle. To find the first angle of the triangle i had to go back a step and make an angle between the two radii. Making another right triangle, i found that it was 22, subtracted it from 180 giving me 158 degrees. Since the hypotenuse line from the center of the circle bisected the angle, to find my first angle of the triangle i divided the angle by 2 which gave me 79 degrees. and subtracting 79 and 90 from 180, i found the third angle, which was 11 degrees. And of course, the radius side
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Student 38
Student 39
Student 40
Student 41 was already known, 65. Using law of tangent, i found the hypotenuse which was 340 and through the pythagorean theorem i found the third side and checked my answer. The hypotenuse gave me the distance from
Ms Nomial to the center of the carousal, i needed her distance from the closest part of it. So i simply subtracted 65 (the radius length) from her distance to get her distance from the edge of the carousal.
Since that each point represented 10 feet and i went on and counted up to where Mrs. Nominal would be at on the x-axis which was to the
45 point and just estimated where she'd be.
She would be 50 feet away, for the extra i'd be impossible for mrs. nominal to see them both.
I think Nomaial is about 66 feet from the carousel. my final answer was
42.18
She is 58.5235 ft. from the closest point of the carousel.
First i found the slopes of the radius segments. Then i found the line of intersection of the two lines of site. Then i found the distance from Mrs. Nomial to the carousel.
First I found the equations of Mrs. Nomial's lines of sight using the negative reciprocal of the slopes of both radiuses to the children. Using those line equations, I found the points of intersection to find where Mrs. Nomial was standing. Then I calculated her distance from the center (0,0) and subtracted from it the distance from the center to the outer edge (65). That number was my final answer
I forgot to add the picture.
I guess I read the question wrong the first time, but now, when I needed to calculate how far Mrs. Nomial was from the carousel,
I first made a triangle from where she was to the closest point of the carousel. (Look at the attachment.) Then what I did was find out the distances, plug them into the Pythagorean Theorem since my triangle was a right triangle. Then when I added 55 squared and 20 squared, I got 3425. Then I found the square root of that and got
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Student 42
Student 43
Mrs. Nomial is standing 66.53 feet from the carousel
Mrs. Nomial is standing 129.7 feet from the center of the carousel.
58.5235 feet. distance formula
I noticed that the points of Mrs. Nomial to her daughter at 25, 60 to the center of the carousel creates a right triangle because the definition of a perpendicular line says that it is 90 degrees. I can then find the length of the sides of this triangle. I know that this distance from the center of the carousel to the first daughter at 25,60 is 65 feet. This is because the diameter of the carousel is 130 feet, and so the distance to the both daughters from the center is 65 feet. The daughters have two congruent triangles that share the side formed from the center of the carousel to their mother. Also, since they are tangent segments drawn from the same point (Mrs. Nomial), the distance from each daughter to Mrs. Nomial is the same.
So we have the length of one side. To find the length of the side connecting Mrs. Nomial and her daughter, I used the distance formula. I looked at the grid and found that Mrs. Nomial's coordinates are about 130,20 and it was given that one of her daughter's coordinates is 25,60. So I used these numbers in the distance formula:
d= sqrt(x2-x1)^2 + (y2-y1)^2
= sprt(130-25)^2 + (20-60)^2
=sqrt(11025) + (1600)
=sqrt(12625)
=112.36 ft
Now I have the lengths of two sides of two congruent right triangles so I can use the pythagorean theorem to find the missing side that is common to both.
a^2 + b^2 = c^2
(65)^2 + 112.36)^2 = c^2
4225 + 12624.7696 = c^2
16849.7696 = c^2
sqrt(16849.7696) = c
129.8 ft= c
This shows that the distance from Mrs. Nomial to the center of the carousel is 129.8 ft. You could also say that she is 64.8 ft away from the edge of the carousel, but I don't think that is important.
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Student 44
Student 45
The length of h is
131.52 so the the diamiter is 130 right.
Half of the diamiter is the radius= 65. using distance formula I got
66.52 as the distance from the edge of the caracell to mis nomial.
Mrs. Nomial is about
66 feet from the edge of the carousel.
I'm pretty sure that my answer is correct, although I think I could have used a simpler strategy. I also could have just used the distance formula to find the distance between Mrs. Nomial and the center by using the 0,0 coordinates. However, I was hoping that I would get a better idea of how to solve this while going the long way. so using distance fourmula its the distance from the center of the caracell to where mis nomial is. so d= sqart(130-0)^2+(0-20)^2 equals (131.52) the distance from mis nomial to the center of the caracell. the distance from the edge to mis nomial is 66.52 (thats the raidius - the answer you got for the problem you solved for the distance formula.so 65-131=66.52.
I saw there were two lines in the problem - one from Mrs. Nomial to each daughter. I got an equation for each line and then figured out where the two lines intersected to see where Mrs. Nomail is standing.
Then I used the distance formula to figure out how far away she is.
To get the equation of the lines, I found the slope of each radius going to the girls. Since one end of each radius is (0,0), the slope is just the y-value of the point where the girl is over the x-value.
So the top radius has a slope of 60/25 which reduces to 12/5. The bottom radius has a slope of -52/39 which reduces by 13 to -4/3.
It said in the problem that the lines are perpendicular to the radiuses, so the slopes of the lines are the negative reciprocals of the slopes of the radiuses. The top line has a slope of -5/12 and the bottom one has a slope of 3/4.
Now I know the slope and a point on the line for each line, so I used the point-slope form to find the equations. y - 60 -5 y - (-52) 3
------ = --- --------- = --- x - 25 12 x - 39 4
12(y - 60) = -5(x - 25) 4(y + 52) = 3(x - 39)
12y - 720 = -5x + 125 4y + 208 = 3x - 117
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Student 46 She was 66 ft away from the carousel
12y + 5x = 845 4y - 3x = -325
This is a system of two equations and two variables. I multiplied the second equation by -3 and then added the two equations to cancel out the y.
-12y + 9x = 975
12y + 5x = 845
---------------
14x = 1820
x = 130
Then I put x = 130 into the second equation to get y.
4y - 3(130) = -325
4y - 390 = -325
4y = 65
y = 16.25
Mrs. Nomial is standing at the coordinates (130, 16.25). That looked right on the picture, so I figured I was on the right track. Now I have to find the distance, so I used the distance formula to find how far (130, 16.25) is from (0,0), the center of the carousel. d = SQR[(130 - 0)^2 + (16.25 - 0)^2]
= SQR[16900 + 264.0625]
= SQR[17164.0625]
= 131.01
So she's standing about 131 feet from the center of the carousel. The carousel is 130 feet across, so its radius is 65 feet. The line from the center to Mrs. Nomial represents a radius and then her distance from the edge of the carousel. I subtracted the radius from the 131 feet and got 131 - 65 = 66.
Mrs. Nomial is standing about 66 feet from the edge of the carousel. find the slopes of the radius segments and then use those slopes to determine the slopes of the lines of sight find an equation for each of Mrs. Nomial's lines of sight find the point of intersection of the two lines of sight
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Student 47 Mrs. Normial is 66 feet away from the carisel.
© 1994-2009 Drexel University http://mathforum.org/pows/ find the distance from Mrs. Nomial to the carousel
First I looked up the description of a tangent line to make sure I know what it is. Then knowing that, I used an equation that consists of several prosecess. The first one is that x=x0+(a)cos(t) and y=y0+(A)sin(t). This is the center of the circle points that you plug in. So now you plug all the values in and next you multiply {x=x0+(a)cos(t) y=y0+(A)sin(t)} times {a cos t and a sin t) pluging in the varibles used in the problem in the propor place like x and y equal to 0. Then you use the quadraric formula and do it like this t=(+)(-) cos {-x (+)(-)y and square root (x^2+y^2-a^2)/
(x^2+y^2) and when you are done you get 66 feet.
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