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Edexcel GCSE Maths Paper 1 Help Thread

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Original post by Habiba12
go on https://www.youtube.com/watch?v=-z4SUypJZxo its a bit annoying but trust me itll stay in your head i heard this once in yr 7 and i still remember it helps with circles and pi


LOOL its funny! & Ty :smile:
I have a test tomorrow I'm in college doing level 2 do you think these questions will be on the test
image.jpgAnyone can help?
Original post by Ranadosa
image.jpgAnyone can help?



Here. Sorry can't explain it - don't have time, going to bed now.
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Original post by Ranadosa
image.jpgAnyone can help?


The size of each angle in an n-sided regular polygon is 180(n2)n\frac{180(n-2)}{n} You can use this to find the size of each angle in the octagon.

Use 180(n2)180(n-2) to find the sum of angles in the hexagon.

And the size of the unknown angles in the hexagon will be equal, so you can work that out by dividing (the sum of angles in a hexagon - 280) by 4.
(edited 7 years ago)
Original post by Ranadosa
image.jpgAnyone can help?


In regular Octagon
Ex.Angle = 360/8 = 45 degrees
Int.Angle- 180-45 = 135 degrees

JK is line of symmetry so draw line of symmetr to divide hexagon into two trapeziums. Now the angle 140 degrees has been split into two so now angle JKF and angle KJF are each equal to 70 degrees.

In a trapezium, there a two paralell lines therefore interior angles can add up to 180 degrees. 180-70= 110 degrees

Now take angle inside trapezium away from interior angle of octagon:
135- 110 = 25 degrees
how do you work out the surface area of a solid hemisphere?
Original post by Yasminxb
how do you work out the surface area of a solid hemisphere?


Half of the surface area of a sphere + area of a circle

A=2πr2+2πrA = 2 \pi r^2 + 2 \pi r
Reply 68
Original post by Yasminxb
how do you work out the surface area of a solid hemisphere?


Surface area of a whole sphere is 4πr2 4\pi r^2 . So half of that outer surface would be 2πr2 2\pi r^2 but you have to remember because you 'cut' the sphere in half there is a circle face at the bottom of the hemisphere so we have to add on πr2 \pi r^2 .
So as the other user said the total surface area of a hemisphere is 2πr2+πr2=3πr2 2\pi r^2 + \pi r^2 = 3\pi r^2 .
Original post by TheOtherSide.
Half of the surface area of a sphere + area of a circle

A=2πr2+2πrA = 2 \pi r^2 + 2 \pi r


Original post by B_9710
Surface area of a whole sphere is 4πr2 4\pi r^2 . So half of that outer surface would be 2πr2 2\pi r^2 but you have to remember because you 'cut' the sphere in half there is a circle face at the bottom of the hemisphere so we have to add on πr2 \pi r^2 .
So as the other user said the total surface area of a hemisphere is 2πr2+πr2=3πr2 2\pi r^2 + \pi r^2 = 3\pi r^2 .


thank youu :biggrin::heart:
Right its hard to explain but you need to find the midpoint of the current line first. Then this will give you an x and y co-ordinate which you use later on. You then have to find the gradient from the current line, it will either be in an equation you are given or you will have to work it out from the diagram given (height÷length). This gradient will probably be in a fraction form, you flip this fraction to find the gradient of the perpendicular line. Then tou have to substitute everything you have found out into the equation y=mx+c...so... you would take the x and y from the midpoint and the gradient (m). Then you just need to make c the subject of your equation.then finally substitue the gradient and c into your equation y=mx+c to get your new equation for the perpendicular line.... sooo recap.
-find the midpoint xy
-find the gradient of current line (ab)
-find the gradient of new line (ab flipped)
-substitue the gradient and x and y into y=mx+c
-then find c and substitue one last time to get you answer

I hope this made some sense... good luck!


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Original post by Jaff_
Yo! Can someone explain the alternate segment therom and proof about consecutive numbers. Thanks, much appreciated!


Always use n and n+1 to show consecutive numbers, there was one a few years back about the difference of two squares of consecutive numbers and it was a case of multiplying n squared and n+1 squared and collecting like terms, sorry if it's vague
Good luck!
L
Reply 72
Yo guys I did my IGCSE Maths a year early in 2015 June, got 97% lol. Anyways, reply to this or PM me if u got any questions, free and happy to help (if I can) lol!

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