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Atomic Boo

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Teaching standards are definately going down. One of my maths teachers teaches directly from the textbook, and if a student doesn't understand something, she can't help them because she doesn't understand it (this is C2 we're talking about, definately not the hardest thing in the world). Having said that, one of my maths teachers is a genious, firsts at Cambridge, I envy him.

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Could you teach me how to get an A* at GCSE

 

@Chuck: Where do you live? If its anywhere near Oxford, £10 an hour (cheap for private tuition). I'll get you an A*. no joke.

 

@Fruit gun: S-Cool And Wikipedia can help. Local Libraries are also good for GCSE level work. Start by planning, A well done plan helps all essays and coursework. My GCSE project was on urban development in Doncaster so i can't help that much.

 

EDIT: *looks at location under chucks name. sighs*. Anyone else?

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Teaching standards are definately going down. One of my maths teachers teaches directly from the textbook, and if a student doesn't understand something, she can't help them because she doesn't understand it (this is C2 we're talking about, definately not the hardest thing in the world). Having said that, one of my maths teachers is a genious, firsts at Cambridge, I envy him.

Haha, C2 should be ridiculously easy for a maths teacher.

 

One of my maths teachers is a bit of a genius. He actually did an entire FP2 paper on the board in 5 minutes.

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@Chuck: Where do you live? If its anywhere near Oxford, £10 an hour (cheap for private tuition). I'll get you an A*. no joke.

 

@Fruit gun: S-Cool And Wikipedia can help. Local Libraries are also good for GCSE level work. Start by planning, A well done plan helps all essays and coursework. My GCSE project was on urban development in Doncaster so i can't help that much.

 

EDIT: *looks at location under chucks name. sighs*. Anyone else?

 

Damn you, stop stealing my market. Although i'm more of a backstreet version of Chris. I can't guarantee that I'll get you an A*. In fact, I would probably numb your brain.

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Has anyone got any Advice for me. As I am doing AS Maths and out of my 4 AS's its the one im stuggling with,(I blame my absolutely useless Maths teacher). I've checked the maths tutors in my immediate area and none of them do above GCSE. :hmm: I really need an A in Maths to strenghten my pursuit of doing medicine at Uni. No amount of work I put into Maths seems to help me. (My own Maths teacher said Im useless and that he dislikes me:weep: ) I think it may have affected my confidence in my own ability.

 

any Advice?:hmm:

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Well first of all ignore the tosser. Any teacher that says that to a student is obviously a shit teacher. I'm only at GCSE myself so I can't help you really. But try searching the internet. You might find something that helps.

 

Cheers mate. Will do: peace:

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Physics question!

 

I'm pretty sure I know this, since it's pretty basic - but I just want to make sure.

 

In a circuit, when the voltage increases - the current does too, but the resistance stays the same. Correct? And in this same circuit, when the voltage is increased the power is increased too?

 

Also, why is the total resistance in a parallel circuit generally lower than the resistance that is actually there, if that makes sense.

 

Thanks

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Ok, handy formula. V=IR

 

Resistance ® is a property of the material and so (to first approximation) It will stay the same if voltage (V) is increases. Current (I) will increase proportionally to Voltage. (This isn't exactly true as higher current creates a heating effect, whihc will increase resistance, but probably best to ignore that)

 

Power = V times I = I^2 times R

 

so increase in voltage will increase power.

 

For the resistance question think about it a little like a road with speed bumps. Having one road (series) with a single speed bump ®, and cars pass slowely over the speed bump. Having a Speed bump twice as high, means the cars must slow down twice as much (2R). Having two roads side by side (paralllel) each having one speed bump. Even though the cars have to slow down the same amount as the oigional road, because there are two roads twice as many cars pass, so the effective resistance is halved (R/2).

 

Thats a really dodgy analogy but its the best i can do atm.

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Has anyone got any Advice for me. As I am doing AS Maths and out of my 4 AS's its the one im stuggling with,(I blame my absolutely useless Maths teacher). I've checked the maths tutors in my immediate area and none of them do above GCSE. :hmm: I really need an A in Maths to strenghten my pursuit of doing medicine at Uni. No amount of work I put into Maths seems to help me. (My own Maths teacher said Im useless and that he dislikes me:weep: ) I think it may have affected my confidence in my own ability.

 

any Advice?:hmm:

 

Have you got any chance to purchase AS level maths like helpbooks at your college, we do at ours and so far I used it for the C1 test and aced it man! seriously if its a big blue book, it helps :D Other than that, see if you can find a private tutor, 1-1 can make a difference I reckon (not from personal experience sorry).

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Ok, handy formula. V=IR

 

Resistance ® is a property of the material and so (to first approximation) It will stay the same if voltage (V) is increases. Current (I) will increase proportionally to Voltage. (This isn't exactly true as higher current creates a heating effect, whihc will increase resistance, but probably best to ignore that)

 

Power = V times I = I^2 times R

 

so increase in voltage will increase power.

 

For the resistance question think about it a little like a road with speed bumps. Having one road (series) with a single speed bump ®, and cars pass slowely over the speed bump. Having a Speed bump twice as high, means the cars must slow down twice as much (2R). Having two roads side by side (paralllel) each having one speed bump. Even though the cars have to slow down the same amount as the oigional road, because there are two roads twice as many cars pass, so the effective resistance is halved (R/2).

 

Thats a really dodgy analogy but its the best i can do atm.

My preferred analogy is to think of it as water flowing through pipes, with the width of the pipe being resistance. Water will flow faster through two narrow pipes at once than through one narrow pipe.

 

(I know this analogy is a bit flawed, but that's what makes it an analogy...)

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Have you got any chance to purchase AS level maths like helpbooks at your college, we do at ours and so far I used it for the C1 test and aced it man! seriously if its a big blue book, it helps :D Other than that, see if you can find a private tutor, 1-1 can make a difference I reckon (not from personal experience sorry).

 

You are a funking LEGEND! I found out we do stock those books at my school and have bought and started using them.

 

Thanx dude!:grin:

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Ok, handy formula. V=IR

 

Resistance ® is a property of the material and so (to first approximation) It will stay the same if voltage (V) is increases. Current (I) will increase proportionally to Voltage. (This isn't exactly true as higher current creates a heating effect, whihc will increase resistance, but probably best to ignore that)

 

Power = V times I = I^2 times R

 

so increase in voltage will increase power.

 

For the resistance question think about it a little like a road with speed bumps. Having one road (series) with a single speed bump ®, and cars pass slowely over the speed bump. Having a Speed bump twice as high, means the cars must slow down twice as much (2R). Having two roads side by side (paralllel) each having one speed bump. Even though the cars have to slow down the same amount as the oigional road, because there are two roads twice as many cars pass, so the effective resistance is halved (R/2).

 

Thats a really dodgy analogy but its the best i can do atm.

 

Thanks a lot! You really helped!

 

edit: nevermind

 

Bump for maths help needed.

 

I get given 2 functions.

 

y = 4 - x

y = x^2 + 2x

 

To find where they intersect I just make 'em equal to each correct?

 

So I'd get

 

x^2 + 3x - 4 = 0

 

Then work that out?

 

Yeah, first work it out

 

(x - 1)(x + 4) = 0

 

Then see the two values for x.

 

x = 1 or x = -4

 

Then using that, put it into one of your y equations.

 

y = 4 - x

y = 4 - 1

y = 3

 

And the other x value

 

y = 4 - x

y = 8

 

So they intersect at points

(1, 3) and (-4, 8)

 

That may make no sense... sorry. :heh:

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Question;

 

If log32 = b express log318 in terms of b

 

I dont mind logarithms, it's just I always foget the method for this one and I left my books in school. :p

 

It's for three marks on an additional GCSE test, which is out of 100, so i'm hesitant to say the answer is 9b =/

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Guest Stefkov

The gradient of a curve is given dy/dx = 2x^-½, and the curve passes through the point (4,5). Find the equation of the curve.

 

I don't know why I'm resitting the maths C1, i did crap first time I will do again. I can't even remember how to do this question..I feel so beat.

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1-up Mushroom

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