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 11-24-2009, 11:20 PM #1 Junior Member   Join Date: Apr 2009 Posts: 58 ACT math need help!!! If anyone have the green ACT booklet that the school gives out and can help me with these problems would be great. # 37, 41, 50, 52, and 53. Most of the problems has picture which is why I didn't posted up the question Reply
 11-24-2009, 11:44 PM #2 Member   Join Date: Jul 2009 Posts: 612 Thats really vague, Use paint and the drawing tools to make the pictures and write up the questions, then itll be much easier for you to get help. Reply
 11-24-2009, 11:55 PM #3 Junior Member   Join Date: Apr 2009 Posts: 58 37. As shown in the standard (x,y) coordinate plane below, P(6,6) lies on the circle with center (2,3) and the radius 5 coordinate units. What are the coordinates of the image of P after the circle is rotated 90 degree clockwise about the center of the circle. A(2,3) B(3,2) C (5,-1) D(6,0) E (7,3) http://img204.imageshack.us/img204/1849/87328288.png Reply
 11-25-2009, 10:15 AM #4 Member   Join Date: Sep 2009 Location: UNC-CH Posts: 544 What's the answer? Reply
 11-25-2009, 10:26 AM #5 Member   Join Date: Sep 2009 Location: UNC-CH Posts: 544 I'm assuming that your sketch is drawn like in the problem. So when you rotating the circle 90 degrees upon (2,3), it seems as though P is the fourth quadrant which has the coordinates (x, -y). C is the only one that's like that. Also, A and B just don't make sense. Please correct me if I'm wrong. Reply
 11-25-2009, 10:29 AM #6 New Member   Join Date: Nov 2009 Posts: 15 You're right C is the correct answer. Reply
 11-25-2009, 12:31 PM #7 Junior Member   Join Date: Apr 2009 Posts: 58 Yes C is the correct answer ,but the different between C and D are so close -1 and 0. I did not know whether 90 degree would pass the x-axis or not...It seem like an estimating problem??? Anyone else>??? Reply
 11-25-2009, 01:02 PM #8 Member   Join Date: Sep 2009 Location: UNC-CH Posts: 544 Estimating would be the easiest way to go about doing this problem. You could, however, try to make a right triangle and find the distances but that's just too much work. I don't know if its worth the time, or if it is even feasible. Reply
 11-25-2009, 02:20 PM #9 Member   Join Date: Jul 2009 Posts: 612 ^ That's what I went ahead and did, it turned into an AP Algebra problem. Estimating is the way to go Reply
 11-25-2009, 02:26 PM #10 Junior Member   Join Date: Apr 2009 Posts: 58 cowking15: If you dont mind can you show me the math way to do this problem? Reply
 11-25-2009, 04:33 PM #11 Junior Member   Join Date: Nov 2009 Posts: 31 Does that green book have an exam number, like 61d, 58b, etc.? I got a few exams from teachers but they did not come in a green book, might be able to help you out... Reply
 11-25-2009, 07:43 PM #12 Junior Member   Join Date: Apr 2009 Posts: 58 ACT-64E practice Reply
 11-25-2009, 07:50 PM #13 Member   Join Date: Jul 2009 Posts: 612 Ducky, I didnt wanna fully work it out but this is the process. Since when rotation an point, the distance from the center stays the same to the point right? SO you know that the distance from the center to the point equals the distance from the center to the new point you use the center coordinates 2,3 the orig coors 6,6 and the new coords x,y to solve it. It gets pretty hectic Reply
 03-31-2012, 08:27 PM #14 New Member   Join Date: Mar 2012 Posts: 1 You could solve this problem in an easy way like the following : 1-Draw a line from the point (6,6) to the other side of the circle witch will make a 180 degree. 2- Because the circle is rotated clockwise you will draw a line from the midpoint of the line"vertical line 90 degree" you draw in step one and will go in the other side of the circle witch is toward the clock rotate. 3-when you connect the second line with the circle you will find that the circle is mating with the line in the point (5,-1). Reply

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