Where can I find challenging Geometry problems?
Q. Dear readers, I need to find some very challenging geometry problems based on the point-line distance formula, stewart's theorem, ptolemy's theorem, mass points, or the inradius/circumradius formulas. The problems can also include whatever other theorems students in high school taking honors geometry should have learned. Can anyone please give me a website that gives problems like this and with detailed explanations (worked out solutions)? Thanks.
Asked by ERL - Wed May 23 19:38:25 2007 - - 3 Answers - 0 Comments
A. try typing in challenging geometry problems on google
Answered by jimbothe_smartguy - Wed May 23 19:43:12 2007
Q. Dear readers, I need to find some very challenging geometry problems based on the point-line distance formula, stewart's theorem, ptolemy's theorem, mass points, or the inradius/circumradius formulas. The problems can also include whatever other theorems students in high school taking honors geometry should have learned. Can anyone please give me a website that gives problems like this and with detailed explanations (worked out solutions)? Thanks.
Asked by ERL - Wed May 23 19:38:25 2007 - - 3 Answers - 0 Comments
A. try typing in challenging geometry problems on google
Answered by jimbothe_smartguy - Wed May 23 19:43:12 2007
If the service menu is of no help, how does a technician fix geometry problems on a television?
Q. I have a crt hdtv with noticeable geometry distortions. I have tried everything in the service menu to fix them, but nothing has worked. Assuming the problems are fixable and the service menu is of no help, what would a technician do to correct them? And is it practical to consider having this work done? Thanks. The tv is a Samsung TX-p2675wh. The geomwtry problems are basically vertical lines are not straight from the bottom to the top of the screen. When they get to the top, they sort of bend off to the right. It's noticeable when a person is in front of a doorway, and the frame looks all crooked.
Asked by Danny27 - Fri Apr 20 19:06:12 2007 - - 2 Answers - 0 Comments
A. They can always replace the video board and see if that helps.
Answered by TV guy - Fri Apr 20 19:31:42 2007
Q. I have a crt hdtv with noticeable geometry distortions. I have tried everything in the service menu to fix them, but nothing has worked. Assuming the problems are fixable and the service menu is of no help, what would a technician do to correct them? And is it practical to consider having this work done? Thanks. The tv is a Samsung TX-p2675wh. The geomwtry problems are basically vertical lines are not straight from the bottom to the top of the screen. When they get to the top, they sort of bend off to the right. It's noticeable when a person is in front of a doorway, and the frame looks all crooked.
Asked by Danny27 - Fri Apr 20 19:06:12 2007 - - 2 Answers - 0 Comments
A. They can always replace the video board and see if that helps.
Answered by TV guy - Fri Apr 20 19:31:42 2007
geometry PROBLEMS!!!?
Q. The surface area of a cube is 864 square inches. FInd the length of the lateral edge. ??? confusing. wouldn't you divide by 6 and then divide by 4 to get the length of each side of the square first? I ended up getting 36 for each side with an area of 2592 for thef 2b part of the equation. so I have 864=144h+2592 which i know is wayyy wrong? I'm so confused on what to do.
Asked by ??? ? - Tue Dec 8 15:36:25 2009 - - 1 Answers - 0 Comments
A. A cube is a three-dimensional square, indicating all sides are equal. Dividing by six gives you 144 square inches, so it is obviously 12 x 12 inches, ne?
Answered by Bert - Tue Dec 8 15:45:07 2009
Q. The surface area of a cube is 864 square inches. FInd the length of the lateral edge. ??? confusing. wouldn't you divide by 6 and then divide by 4 to get the length of each side of the square first? I ended up getting 36 for each side with an area of 2592 for thef 2b part of the equation. so I have 864=144h+2592 which i know is wayyy wrong? I'm so confused on what to do.
Asked by ??? ? - Tue Dec 8 15:36:25 2009 - - 1 Answers - 0 Comments
A. A cube is a three-dimensional square, indicating all sides are equal. Dividing by six gives you 144 square inches, so it is obviously 12 x 12 inches, ne?
Answered by Bert - Tue Dec 8 15:45:07 2009
How do you solve geometry problems like this?
Q. This is just an example, I have no idea if there is an actual answer to this problem: If a chord is 9 feet long and is 3 feet away from the center of the circle, how big is the circle? I think the problems are phrased something like that. Is there a formula for that?
Asked by KenF - Sat Feb 27 20:43:28 2010 - - 3 Answers - 0 Comments
A. From what you just told me I can tell the radius of the circle is 12ft. There is a formula it is A=pi x r x r You know the Radius( the length from the center to the outside of the circle) is 12 Pi is apromixmetly 3.14 so: Area=3.14 x 12 x 12 Area=452.16
Answered by grasseater - Sat Feb 27 20:54:58 2010
Q. This is just an example, I have no idea if there is an actual answer to this problem: If a chord is 9 feet long and is 3 feet away from the center of the circle, how big is the circle? I think the problems are phrased something like that. Is there a formula for that?
Asked by KenF - Sat Feb 27 20:43:28 2010 - - 3 Answers - 0 Comments
A. From what you just told me I can tell the radius of the circle is 12ft. There is a formula it is A=pi x r x r You know the Radius( the length from the center to the outside of the circle) is 12 Pi is apromixmetly 3.14 so: Area=3.14 x 12 x 12 Area=452.16
Answered by grasseater - Sat Feb 27 20:54:58 2010
Can any of you teach me how to do these geometry problems?
Q. Ok here is what i need to learn for geometry was there for a week and missed the lessons. Finding the length of a line on a coordinate plane Finding the distance between two points Finding coordinates of the mid point. Any help would be awesome.
Asked by The angry Matt - Thu Sep 24 13:48:43 2009 - - 2 Answers - 0 Comments
A. So let's say you have a line (segment) in the coordinate plane. It goes from (a,b) to (c,d). Let's say, as an example, (0,3) to (3,7). The midpoint of the line is easy: you just average the X coordinates and the Y coordinates. So, the midpoint is ((a+c)/2, (b+d)/2), in this case, (1.5,5). For the length of the line, the easiest thing is to use the Pythagorean theorem. You can make a triangle with the line as the hypotenuse, where one leg is horizontal and the other is vertical. The lengths of the two legs are the differences between the coordinates: |a-c| and |b-d|, in this case, 3 and 4. So the length is SQRT(|a-c|^2 + |b-d|^2), in this case, SQRT(3^2 + 4^2) = SQRT(25) = 5. The distance between two points is the same as the… [cont.]
Answered by Cola - Thu Sep 24 13:59:14 2009
Q. Ok here is what i need to learn for geometry was there for a week and missed the lessons. Finding the length of a line on a coordinate plane Finding the distance between two points Finding coordinates of the mid point. Any help would be awesome.
Asked by The angry Matt - Thu Sep 24 13:48:43 2009 - - 2 Answers - 0 Comments
A. So let's say you have a line (segment) in the coordinate plane. It goes from (a,b) to (c,d). Let's say, as an example, (0,3) to (3,7). The midpoint of the line is easy: you just average the X coordinates and the Y coordinates. So, the midpoint is ((a+c)/2, (b+d)/2), in this case, (1.5,5). For the length of the line, the easiest thing is to use the Pythagorean theorem. You can make a triangle with the line as the hypotenuse, where one leg is horizontal and the other is vertical. The lengths of the two legs are the differences between the coordinates: |a-c| and |b-d|, in this case, 3 and 4. So the length is SQRT(|a-c|^2 + |b-d|^2), in this case, SQRT(3^2 + 4^2) = SQRT(25) = 5. The distance between two points is the same as the… [cont.]
Answered by Cola - Thu Sep 24 13:59:14 2009
I need help with proofs and other geometry problems?
Q. Can you guys give me any sites where i can get Geometry help online for free. And any sites where i can get online help for proof statements.
Asked by cool - Wed Dec 12 22:12:44 2007 - - 1 Answers - 0 Comments
A. try math.com i think it has proof stuff or ask your teacher or other teachers about sites like that
Answered by steven l - Wed Dec 12 22:20:56 2007
Q. Can you guys give me any sites where i can get Geometry help online for free. And any sites where i can get online help for proof statements.
Asked by cool - Wed Dec 12 22:12:44 2007 - - 1 Answers - 0 Comments
A. try math.com i think it has proof stuff or ask your teacher or other teachers about sites like that
Answered by steven l - Wed Dec 12 22:20:56 2007
How would I answer these geometry problems?
Q. I don't understand, please help :// thank you!! 1. Two supplementary angles have measures in the ratio 3:7. Find the measure of the larger angle. 2. The perimeter of a triangle is 48 cm and the lengths of the sides are in a ratio of 3:4:5. Find the length of the shorter side. I know the answers are 126 degrees and 12 cm, I just don't understand how to get those :/
Asked by ><>MaY<>< - Mon Jan 25 17:37:25 2010 - - 1 Answers - 0 Comments
A. 1) Supplement angles add up to 180 so you can just change the numbers in ratio to x's. The equation would be 3x + 7x = 180. Then solve for x to get x = 18. 7x would obviously be the bigger angle so you can plug in 18 to get 7(18) = 126. 2) You know the perimeter of the triangle. Like the first question, change the 3:4:5 to 3x, 4x, and 5x. The equation would be 3x + 4x + 5x = 48. Then solve for x to get x = 4. The shortest side would be the 3x so you can plug in 4 to get 3(4) = 12. Hope it helps :)
Answered by Wendy - Mon Jan 25 17:47:50 2010
Q. I don't understand, please help :// thank you!! 1. Two supplementary angles have measures in the ratio 3:7. Find the measure of the larger angle. 2. The perimeter of a triangle is 48 cm and the lengths of the sides are in a ratio of 3:4:5. Find the length of the shorter side. I know the answers are 126 degrees and 12 cm, I just don't understand how to get those :/
Asked by ><>MaY<>< - Mon Jan 25 17:37:25 2010 - - 1 Answers - 0 Comments
A. 1) Supplement angles add up to 180 so you can just change the numbers in ratio to x's. The equation would be 3x + 7x = 180. Then solve for x to get x = 18. 7x would obviously be the bigger angle so you can plug in 18 to get 7(18) = 126. 2) You know the perimeter of the triangle. Like the first question, change the 3:4:5 to 3x, 4x, and 5x. The equation would be 3x + 4x + 5x = 48. Then solve for x to get x = 4. The shortest side would be the 3x so you can plug in 4 to get 3(4) = 12. Hope it helps :)
Answered by Wendy - Mon Jan 25 17:47:50 2010
Can someone help me solve some geometry problems?
Q. 1. Prove that in an equilateral triangle if a line segment is a median, then it is an altitude. 2. Ste the converse proposition to exercise 1 for sn equilateral triangle, Do you think it is true? Can you prove it?
Asked by caramel - Mon Jan 4 22:14:00 2010 - - 1 Answers - 0 Comments
A. hint : the line for the median { altitude } will generate two congruent triangles { corresponding sides have same length }...you go from there
Answered by ted s - Mon Jan 4 23:13:53 2010
Q. 1. Prove that in an equilateral triangle if a line segment is a median, then it is an altitude. 2. Ste the converse proposition to exercise 1 for sn equilateral triangle, Do you think it is true? Can you prove it?
Asked by caramel - Mon Jan 4 22:14:00 2010 - - 1 Answers - 0 Comments
A. hint : the line for the median { altitude } will generate two congruent triangles { corresponding sides have same length }...you go from there
Answered by ted s - Mon Jan 4 23:13:53 2010
Could anyone please help me with these 2 simple geometry problems?
Q. I understand these might be simple but I'm having difficulties understanding them as I don't have a book handy yet and just started with the course. For problem A I need to find the value of angle A. For problem B I need to find the x,y and polar coordinates for both points A and B. Please help me out here. If you don't wish to tell me the answer then at least name the theorem or postulate that I need to solve this. Thank you.
Asked by Elvis - Fri Aug 14 22:55:33 2009 - - 1 Answers - 0 Comments
A. Where are Problem A and B?
Answered by Msean - Sat Aug 15 13:49:41 2009
Q. I understand these might be simple but I'm having difficulties understanding them as I don't have a book handy yet and just started with the course. For problem A I need to find the value of angle A. For problem B I need to find the x,y and polar coordinates for both points A and B. Please help me out here. If you don't wish to tell me the answer then at least name the theorem or postulate that I need to solve this. Thank you.
Asked by Elvis - Fri Aug 14 22:55:33 2009 - - 1 Answers - 0 Comments
A. Where are Problem A and B?
Answered by Msean - Sat Aug 15 13:49:41 2009
Please help me with a few of these geometry problems, 10 points to best answer ?
Q. 1. What is the sum of the measures of the exterior angles, one at each vertex, of an octagon? 2. The sum of the measures of the angles of a regular polygon is 4500. How many sides does the polygon have? 3. What is the measure of an exterior angle of a regular polygon with 36 sides?
Asked by Anonymous - Thu Dec 3 12:39:50 2009 - - 1 Answers - 0 Comments
A. 1. the formula for the interior angle of a regular n-gon is 180(n-2)/n so for an octogon it'd be 180(8-2)/8 = 135 the exterior angle is 360 less the interior angle, so its 225. multiply this by 8 for each of the vertices to get 1800 and there's your answer! 2. for the sum of the interior angles just don't divide by n in the formula above so you have 180(n-2). but you know what the sum is so you have the equation 180(n-2) = 4500 n-2 = 25 n = 27 the polygon has 27 sides. 3. same again using the first formula 180(36-2)/36 = 170 170 is the interior angle, so 190 is the exterior one.
Answered by isla - Thu Dec 3 13:08:22 2009
Q. 1. What is the sum of the measures of the exterior angles, one at each vertex, of an octagon? 2. The sum of the measures of the angles of a regular polygon is 4500. How many sides does the polygon have? 3. What is the measure of an exterior angle of a regular polygon with 36 sides?
Asked by Anonymous - Thu Dec 3 12:39:50 2009 - - 1 Answers - 0 Comments
A. 1. the formula for the interior angle of a regular n-gon is 180(n-2)/n so for an octogon it'd be 180(8-2)/8 = 135 the exterior angle is 360 less the interior angle, so its 225. multiply this by 8 for each of the vertices to get 1800 and there's your answer! 2. for the sum of the interior angles just don't divide by n in the formula above so you have 180(n-2). but you know what the sum is so you have the equation 180(n-2) = 4500 n-2 = 25 n = 27 the polygon has 27 sides. 3. same again using the first formula 180(36-2)/36 = 170 170 is the interior angle, so 190 is the exterior one.
Answered by isla - Thu Dec 3 13:08:22 2009
i need answers that are correct to these Geometry problems?
Q. 1. What's the perimeter of a regular hexagon whose sides are 12.7 yards long? 2. If RX= 2y+2x , TX=3y-1 , and SV=28 what's the values of x+y?
Asked by bettyboopboop - Mon May 11 12:15:12 2009 - - 1 Answers - 0 Comments
Q. 1. What's the perimeter of a regular hexagon whose sides are 12.7 yards long? 2. If RX= 2y+2x , TX=3y-1 , and SV=28 what's the values of x+y?
Asked by bettyboopboop - Mon May 11 12:15:12 2009 - - 1 Answers - 0 Comments
How do you solve these geometry problems?
Q. 1. The metric system of weights and measures works with multiples of ___. 2. 5.7 centigrams to grams 3. 120 pounds converted to kilograms show work: 4. thirty degrees C to F 5. 15 kilometers to miles 6. 8 yards to meters 7. find the number of liters in 14.3 gallon of gasoline 8. plums cost $0.69 per pound. Find the cost per kilogram. thank you so much!
Asked by teebaby - Sun Mar 29 18:29:51 2009 - - 1 Answers - 0 Comments
Q. 1. The metric system of weights and measures works with multiples of ___. 2. 5.7 centigrams to grams 3. 120 pounds converted to kilograms show work: 4. thirty degrees C to F 5. 15 kilometers to miles 6. 8 yards to meters 7. find the number of liters in 14.3 gallon of gasoline 8. plums cost $0.69 per pound. Find the cost per kilogram. thank you so much!
Asked by teebaby - Sun Mar 29 18:29:51 2009 - - 1 Answers - 0 Comments
Solve geometry problems involving circles, arcs and sectors?
Q. P and Q are two points on the side BC of a triangle ABC such that AP is the bisector of the angle BAC and AQ is the altitude of the triangle. If angle ACB is 40 degrees and ABC is 60 degrees, determine the magnitude of angle PAQ.
Asked by ruhail774 - Fri Nov 27 06:19:38 2009 - - 3 Answers - 0 Comments
Q. P and Q are two points on the side BC of a triangle ABC such that AP is the bisector of the angle BAC and AQ is the altitude of the triangle. If angle ACB is 40 degrees and ABC is 60 degrees, determine the magnitude of angle PAQ.
Asked by ruhail774 - Fri Nov 27 06:19:38 2009 - - 3 Answers - 0 Comments
How to solve Algebra/Geometry short problems?
Q. Please leave an explanation on how you solved the problems: What is the circumference of a circle inside of which is inscribed a triangle with side lengths 3, 4, and 5? Point P is inside regular octagon abcdefgh so that triangle ABP is equilateral. How many degrees are in angle APC?
Asked by partay*chicka - Sun Mar 7 19:01:42 2010 - - 1 Answers - 0 Comments
Q. Please leave an explanation on how you solved the problems: What is the circumference of a circle inside of which is inscribed a triangle with side lengths 3, 4, and 5? Point P is inside regular octagon abcdefgh so that triangle ABP is equilateral. How many degrees are in angle APC?
Asked by partay*chicka - Sun Mar 7 19:01:42 2010 - - 1 Answers - 0 Comments
Can you please answer these geometry math problems for me?
Q. Here are the links to the problems: Problem 1: Problem 2: I need these both answered by tonight, i have tried solving them for awhile now, but I am terrible at math. Could someone please help me? Sorry about the other picture of Problem 1, I did not realize it was incomplete, here is a new link for it Here are the answers I got for problem 1: A: 12.5 B:20.519 C: 25 D: 39.269 Are these correct? Problem 2: Total area of shaded: 275.685 Total area unshaded: 293.362 So my final answer would be that the player does not have an equal chance.
Asked by mkatesoccer - Thu Feb 12 21:10:38 2009 - - 1 Answers - 0 Comments
A. 1) Two of those are easy,although the picture is horribly incomplete. D is 1/8 of the area of the whole circle, where the circle has radius 10 and hence area 100 pi. C seems to be a 5x5 square, with area 25. From that you can get A+B. Also, A+2B+C is 1/4 the area of the circle, and you know C, so you can get A+2B. Well, if you know A+B and A+2B, you subtract the first from the second to get B. And then you subtract that from A+B to get A.
Answered by Curt Monash - Thu Feb 12 23:07:53 2009
Q. Here are the links to the problems: Problem 1: Problem 2: I need these both answered by tonight, i have tried solving them for awhile now, but I am terrible at math. Could someone please help me? Sorry about the other picture of Problem 1, I did not realize it was incomplete, here is a new link for it Here are the answers I got for problem 1: A: 12.5 B:20.519 C: 25 D: 39.269 Are these correct? Problem 2: Total area of shaded: 275.685 Total area unshaded: 293.362 So my final answer would be that the player does not have an equal chance.
Asked by mkatesoccer - Thu Feb 12 21:10:38 2009 - - 1 Answers - 0 Comments
A. 1) Two of those are easy,although the picture is horribly incomplete. D is 1/8 of the area of the whole circle, where the circle has radius 10 and hence area 100 pi. C seems to be a 5x5 square, with area 25. From that you can get A+B. Also, A+2B+C is 1/4 the area of the circle, and you know C, so you can get A+2B. Well, if you know A+B and A+2B, you subtract the first from the second to get B. And then you subtract that from A+B to get A.
Answered by Curt Monash - Thu Feb 12 23:07:53 2009
How would i do this geometry problems?
Q. Find the distance from G(-3, -1) to the line whose equation is y = 4. Find the distance between the parallel lines m and n whose equations are y = x + 4 and y = x - 6, respectivel
Asked by *wearing my heart on my sleeve* - Thu Jun 4 11:44:18 2009 - - 1 Answers - 0 Comments
A. for second, choose point on one of the lines and do the same. Not sure your math level so dont want to use vectors unless you know them.
Answered by mackler - Thu Jun 4 11:51:19 2009
Q. Find the distance from G(-3, -1) to the line whose equation is y = 4. Find the distance between the parallel lines m and n whose equations are y = x + 4 and y = x - 6, respectivel
Asked by *wearing my heart on my sleeve* - Thu Jun 4 11:44:18 2009 - - 1 Answers - 0 Comments
A. for second, choose point on one of the lines and do the same. Not sure your math level so dont want to use vectors unless you know them.
Answered by mackler - Thu Jun 4 11:51:19 2009
can anyone help with these simple geometry problems?
Q. 1. how would multiplying the circumference of a circle by 2.4 affect the area? 2. How would tripling the apothem of a regular octagon affect the area? 3. How would multiplying one diagnal of a kite by 1/7 affect the area?
Asked by J C - Tue Apr 8 21:42:37 2008 - - 1 Answers - 0 Comments
A. 1. Circumference = 2 Pi x 2 x radius Multiplying circumference by 2.4 multiplies radius by 2.4 also Area of circle = Pi x radius x radius So area is increased by 2.4 x 2.4 or 5.76 times
Answered by MathMan TG - Tue Apr 8 21:48:29 2008
Q. 1. how would multiplying the circumference of a circle by 2.4 affect the area? 2. How would tripling the apothem of a regular octagon affect the area? 3. How would multiplying one diagnal of a kite by 1/7 affect the area?
Asked by J C - Tue Apr 8 21:42:37 2008 - - 1 Answers - 0 Comments
A. 1. Circumference = 2 Pi x 2 x radius Multiplying circumference by 2.4 multiplies radius by 2.4 also Area of circle = Pi x radius x radius So area is increased by 2.4 x 2.4 or 5.76 times
Answered by MathMan TG - Tue Apr 8 21:48:29 2008
Need help with some coordinate geometry problems?
Q. I am a little stuck on how to work out the following problems so if anyone could help me with all or any of these problems that would be great. The questions are: 1. The gradient of the line perpendicular to 3x - 4y + 7 = 0 is: A. 3/4 B. 4/3 C. -4/3 D. 3 E. -4 2. Find the equation of the straight line passing through the point (6, -2) and parallel to the line x+2y-1=0
Asked by Will P - Sun Aug 2 08:17:44 2009 - - 4 Answers - 0 Comments
A. Problem 1: Convert the given equation to y = mx + b, where m is the slope. 3x - 4y + 7 = 0 - 4y = - 3x - 7 (Divide all terms by -4) y = (3/4)x + 7/4 Slope is 3/4. In order for a slope to be perpendicular to 3/4, it must be the negative reciprocal of the said slope, so they result to -1 when multiplied together. Thus, it's -4/3. C is the answer. ~~~ Let's determine first the slope of the given equation. Convert it to y = mx + b. x + 2y - 1 = 0 2y = - x + 1 (Divide all terms by 2) y = (-1/2)x + 1/2 Slope is -1/2. Now, in order for our equation to be parallel with that, our slope must also be -1/2. In forming the equation of a line, use this formula: y - y1 = m (x - x1) Substitute our gathered data in the formula. P (6, -2) m = -1/ [cont.]
Answered by Dr. Lucafont - Sun Aug 2 08:26:19 2009
Q. I am a little stuck on how to work out the following problems so if anyone could help me with all or any of these problems that would be great. The questions are: 1. The gradient of the line perpendicular to 3x - 4y + 7 = 0 is: A. 3/4 B. 4/3 C. -4/3 D. 3 E. -4 2. Find the equation of the straight line passing through the point (6, -2) and parallel to the line x+2y-1=0
Asked by Will P - Sun Aug 2 08:17:44 2009 - - 4 Answers - 0 Comments
A. Problem 1: Convert the given equation to y = mx + b, where m is the slope. 3x - 4y + 7 = 0 - 4y = - 3x - 7 (Divide all terms by -4) y = (3/4)x + 7/4 Slope is 3/4. In order for a slope to be perpendicular to 3/4, it must be the negative reciprocal of the said slope, so they result to -1 when multiplied together. Thus, it's -4/3. C is the answer. ~~~ Let's determine first the slope of the given equation. Convert it to y = mx + b. x + 2y - 1 = 0 2y = - x + 1 (Divide all terms by 2) y = (-1/2)x + 1/2 Slope is -1/2. Now, in order for our equation to be parallel with that, our slope must also be -1/2. In forming the equation of a line, use this formula: y - y1 = m (x - x1) Substitute our gathered data in the formula. P (6, -2) m = -1/ [cont.]
Answered by Dr. Lucafont - Sun Aug 2 08:26:19 2009
please help me with these geometry problems?
Q. 1.For which of the following properties does the relationship "is darker than" hold true? 2. /_1 and /_2 are a linear pair. M/_1=73 degrees. Find M/_2.
Asked by Berry A - Wed Aug 13 13:37:29 2008 - - 1 Answers - 0 Comments
A. 1. what does darkness have to do with geometry? 2. linear means supplementary so M/_2 = 180 - 73 = 107
Answered by tryzub91 - Wed Aug 13 14:08:54 2008
Q. 1.For which of the following properties does the relationship "is darker than" hold true? 2. /_1 and /_2 are a linear pair. M/_1=73 degrees. Find M/_2.
Asked by Berry A - Wed Aug 13 13:37:29 2008 - - 1 Answers - 0 Comments
A. 1. what does darkness have to do with geometry? 2. linear means supplementary so M/_2 = 180 - 73 = 107
Answered by tryzub91 - Wed Aug 13 14:08:54 2008
How can I get the answers to my problems in my geometry book?
Q. The book is called Glencoe Mathematics Geometry. I critically need the answers! I know there are selected answers in the back of the book, but only ever other one. I found a great website, Hot-math, but it costs $50 a year and it is not in my budget. Any other ideas?? It would be greatly appreciated!
Asked by ........ - Mon Feb 9 20:19:44 2009 - - 2 Answers - 0 Comments
A. Well, there is this thing called paying attention in class. I did it in my Geometry class, using the same book, and came out with a 93. It works pretty well. All you have to do is use your brain a little and put some effort into it.
Answered by annoyed... - Tue Feb 10 18:29:24 2009
Q. The book is called Glencoe Mathematics Geometry. I critically need the answers! I know there are selected answers in the back of the book, but only ever other one. I found a great website, Hot-math, but it costs $50 a year and it is not in my budget. Any other ideas?? It would be greatly appreciated!
Asked by ........ - Mon Feb 9 20:19:44 2009 - - 2 Answers - 0 Comments
A. Well, there is this thing called paying attention in class. I did it in my Geometry class, using the same book, and came out with a 93. It works pretty well. All you have to do is use your brain a little and put some effort into it.
Answered by annoyed... - Tue Feb 10 18:29:24 2009
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