help. ive forgoten how to simplify fractions.
29 September 2004, 08:18 PM
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help. ive forgoten how to simplify fractions.
i need to know how to simplify fractions again.
4/9 x 5/8
what do i do?
convert to decimals?
help please
4/9 x 5/8
what do i do?
convert to decimals?
help please
29 September 2004, 09:34 PM
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Originally Posted by TelBoy
20/72, or 10/36, or 5/18
Easy.
Easy.
PG:
1) multiply the top numbers together: 4 x 5 =20
2) multiply the bottom numbers together 9 x 8 = 72
3) find a the largest whole number that will divide into both top and bottom (in this case 4)
4) divide both top and bottom by 4 -> 20/4 = 5 and 72 /4 = 18
5) reassemble: 5/18
so 3/8 x 4/7 would be?
8 x 7 = 56
12/56 (divide top and bottom by 4) = 3/14
"those who can..... teach!" (or not
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29 September 2004, 11:02 PM
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29 September 2004, 11:56 PM
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Originally Posted by judgejules
omfg, makes me not want to have kids!
daddy, can you help me with my homework? arrghhhhh flashbacks, nooo the pain
Jules
daddy, can you help me with my homework? arrghhhhh flashbacks, nooo the pain
Jules
I'm just trying to teach my four year old about impersonal pronouns and why she shouldn't use them when referring to someone in the same room.
The nightmare of grammar school comes thudding back.
Steve.
30 September 2004, 10:55 AM
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Makes you wonder what all those years fookin around with your mates at school were in aid of, doesn't it?
I remember it now though
30 September 2004, 11:24 AM
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Originally Posted by judgejules
omfg, makes me not want to have kids!
daddy, can you help me with my homework? arrghhhhh flashbacks, nooo the pain
Jules
daddy, can you help me with my homework? arrghhhhh flashbacks, nooo the pain
Jules
One of my neighbours kids is doing maths A level. At the minute they are doing calculus. His dad comes round every night with the son and his homework to get me to explain/do it. Last night was inverse trig differentiation. He says that the teacher doesn't even understand it
30 September 2004, 11:30 AM
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For those that want a Thursday morning mental workout, the question was
A movie screen on the front wall in your classroom is 16 feet high and positioned 9 feet above your eye-level. How far away from the front of the room should you sit in order to have the ``best" view ?
A movie screen on the front wall in your classroom is 16 feet high and positioned 9 feet above your eye-level. How far away from the front of the room should you sit in order to have the ``best" view ?
30 September 2004, 11:50 AM
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Depends on your eyesite:
20/20 vision etc, do you wear glasses, what light level in the room, reflectance, resolution of screen.
20/20 vision etc, do you wear glasses, what light level in the room, reflectance, resolution of screen.
30 September 2004, 11:53 AM
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But in the interest of mathmeatics:
screen - Distance
size
30"6.25 feet35"7.3 feet40"8.3 feet45"9.4 feet50"10.4 feet55"11.5 feet60"12.5 feet65"13.5 feet
screen - Distance
size
30"6.25 feet35"7.3 feet40"8.3 feet45"9.4 feet50"10.4 feet55"11.5 feet60"12.5 feet65"13.5 feet
30 September 2004, 12:06 PM
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Feet? Thought they worked in metric these days?
Homework - can't wait for daughter to come home with these type of questions!
Homework - can't wait for daughter to come home with these type of questions!
30 September 2004, 04:36 PM
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The screen is 16 ft high as per question.
If you imagine two triangles. The bottom of each one is x. The side of one is 9 and the side of the other is 25 (A side view cross section of the room.) Your viewing distance from the wall is x.
Let A be the angle between the floor and the hypotenuse of your view to the bottow of the screen.
Let B be the angle between the floor and the hypotenuse of your view to the top of the screen.
Let Z = B-A. Now work out the length of x that gives the largest possible value of Z.
from trig it follows (after about 3 pages) that
z=arctan(25/x)-arctan(9/x)
All you have to do now is differentiate the function of z and set the derivative to zero.
If you imagine two triangles. The bottom of each one is x. The side of one is 9 and the side of the other is 25 (A side view cross section of the room.) Your viewing distance from the wall is x.
Let A be the angle between the floor and the hypotenuse of your view to the bottow of the screen.
Let B be the angle between the floor and the hypotenuse of your view to the top of the screen.
Let Z = B-A. Now work out the length of x that gives the largest possible value of Z.
from trig it follows (after about 3 pages) that
z=arctan(25/x)-arctan(9/x)
All you have to do now is differentiate the function of z and set the derivative to zero.
30 September 2004, 04:46 PM
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Originally Posted by fast bloke
The screen is 16 ft high as per question.
If you imagine two triangles. The bottom of each one is x. The side of one is 9 and the side of the other is 25 (A side view cross section of the room.) Your viewing distance from the wall is x.
Let A be the angle between the floor and the hypotenuse of your view to the bottow of the screen.
Let B be the angle between the floor and the hypotenuse of your view to the top of the screen.
Let Z = B-A. Now work out the length of x that gives the largest possible value of Z.
from trig it follows (after about 3 pages) that
z=arctan(25/x)-arctan(9/x)
All you have to do now is differentiate the function of z and set the derivative to zero.
If you imagine two triangles. The bottom of each one is x. The side of one is 9 and the side of the other is 25 (A side view cross section of the room.) Your viewing distance from the wall is x.
Let A be the angle between the floor and the hypotenuse of your view to the bottow of the screen.
Let B be the angle between the floor and the hypotenuse of your view to the top of the screen.
Let Z = B-A. Now work out the length of x that gives the largest possible value of Z.
from trig it follows (after about 3 pages) that
z=arctan(25/x)-arctan(9/x)
All you have to do now is differentiate the function of z and set the derivative to zero.
Oi, don't spoil it - I am half way through this, been doing it in my spare time today!!!
30 September 2004, 04:56 PM
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Originally Posted by AndyC_772
Cancel top & bottom.
4/9 x 5/8
= 1/9 x 5/2
= 5/18
4/9 x 5/8
= 1/9 x 5/2
= 5/18
Am currently learning what a huge increase in ability is needed to jump from GCSE maths to A-Level Maths
30 September 2004, 04:59 PM
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Originally Posted by ajm
Oi, don't spoil it - I am half way through this, been doing it in my spare time today!!!
Thats the easy bit done for you - you can get straight on to the difficult bit
30 September 2004, 05:04 PM
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I don't recall A-level maths requiring a huge increase in ability as such, but it sure as heck requires a huge increase in effort and number of hours put in. I did further maths too and still remember Sundays spent doing past exam papers *shudder*.
What was worst of all, was our maths teacher had a habit of setting us the paper one week, and then all the questions we'd opted out of doing (ie. couldn't do!!) the next week. The b*stard!
Scarier still was that when I went on to do my engineering degree, we covered the whole of the 2 yr further maths course in the first term and a half - and kept that pace up until the end of the course, by which time I was a gibbering wreck. I actually can't do much in the way of maths at all any more, it's fried that little part of my brain. Thankfully my job (electronic engineer) doesn't require any after all that
What was worst of all, was our maths teacher had a habit of setting us the paper one week, and then all the questions we'd opted out of doing (ie. couldn't do!!) the next week. The b*stard!
Scarier still was that when I went on to do my engineering degree, we covered the whole of the 2 yr further maths course in the first term and a half - and kept that pace up until the end of the course, by which time I was a gibbering wreck. I actually can't do much in the way of maths at all any more, it's fried that little part of my brain. Thankfully my job (electronic engineer) doesn't require any after all that