I am completely shit at Maths, pretty please could someone tell me how you're supposed to work out the following so I can hopefully get a C?
1. Factorise fully (n²-a²)-(n-a)²
2. Rearrange the formula to make a the subject: P=n²+a divided by n+a
3. Explain why (n²-a²)-(n-a)² is always an even integer.
4. 8/8 (square root) can be written in the form 8^k. Find the value of k.
hint
if you can answer Q1, Q3 is easy
please dont tell me that is GCSE standard
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They're from the intermediate paper so it's C level. and some people just don't have an aptitude for Maths.
...
1) you need to multiply out the the second bracket, and then work out how many of each power there are.
2) times both sides by n+a then rearrange so that a is on one side and everything else is on the other
3) um
4) 8 = 8^1 and the square root of 8 is 8^(1/2) and dividing by something is the same as subtracting the power so it's: 8^1 - 8^(1/2) = 8^(1-(1/2))
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Thank you!
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Thank you!
shit
that should be 8^1 / 8^(1/2) = 8^(1-(1/2)) at the end. You're welcome, the most use i've made of my degree since graduating probably.
does ^ mean x nowadays?
what's an integer?
I know I got a C at the time but I'm not sure I would these days (without revising anyway)
^ = to the power of
integer = positive whole number.
I used to love maths
I don't remember any of this
And I was like a couple of marks short of FUll Marks in this subject :) :(
i liked
that you only had to get 25% for a C at GCSE.
hrm
= n² - a² - ((n-a)(n-a))
= n² - a² - (n² - 2an + a²)
= n² - a² - n² + 2an - a²
= -2a² + 2an
= 2a(-a + n)
maybe...
i don't really know...
this works
even if usually it's not a good idea to multiply before you factorise.
Another way of doing it would have been n² - a² = ( n + a )( n - a ) then (n²-a²)-(n-a)² = ( n - a )[( n + a)-( n - a )]
^5
That's how I'd have done it.
yo
question 3
The answer to Q1 is 2a(-a + n).
This can also be written as -2a^2 + 2an.
Each term in this formula is divisible by 2.
Therefore (blah equation) must be an even NUMBER...
...but I don't understand why it should be an INTEGER as it doesn't say n and a have to be whole numbers (which is what integers are, basically). Unless it says they do?
For example if n were 2.5 and a were 1.6, the formula would equal (6.25 - 2.56) - (0.81), which would be 2.88. Which is an even number but not an even integer.
Sure, if n and a are themselves integers, then the formula would be an even integer.
Am I shite?
.
Actually, being an integer is a necessary condition for being even, else every number would be even. I'm pretty sure the Q will have stipulated A&N as integers.
cheers KoC
maths makes me sick to my very core
^thisx1000000
Surely you won't get these question for the intermediate paper.
I was shitting it for my higher GCSE maths exam because the material we were learning in lessons seemed difficult, they gave us the impression that these were standard questions. In reality most of the paper wasn't that tough, it was only the odd question here and there. I had some shots of vodka before the exam for my anxiety, I got a "B" though.
I used to love maths when I was younger (obsessed is probably a more accurate word to use), then it started being all about letters and graphs as opposed to numbers. When I was 8, I did the maths work of the 11 year olds, my whole class was fascinated by this strange alien textbook, it was great I felt so important.