I've got my exams coming up in a few months, and I'm having some trouble with my Maths, so I thought I'd ask some people here how to do some of the calculations
It'd help a lot if I could ask some questions and if someone could answer them and explain how to do them. (Don't worry, you're not helping me cheat, it's revision ). It might help to know that I'm 15, so (obviously), I don't know too much advanced maths.


Proportion
The area of the screen of a television set is A square inches.
The length of the diagonal of the screen is d inches.
A is directly proportional to the square of d.

A television set with an area of 90 square inches has a diagonal length of 15 inches.

a) Find an equation connecting A and d
b} Find the area of the screen of a television set with a diagonal length of 20 inches.
_ _ _ _ _

Inverse proportion
The number of days, D, to complete a project is inversely proportional to the number of poeple, P, who work on the project.
a) The project takes 18 days to complete with 150 people working on it.
Find an equation connecting D and P
_ _ _ _ _

Simplification
a) Simplify [square root]18 + [square root]32
b} Rationalise 1
[square root] 6
_ _ _ _ _

And if someone could explain what SURDS and Integers are, that would also help a lot.

Thanks

Intergers are WHOLE numbers..

For Simplification A

you have SR(square root) 18. find a multiples of 18 to divide by...such as SR3 x SR6...SR2 x SR9....the latter would work because SR9=3
So it would be 3 x SR2. Thats simplified...now you can do the other..

For B

to bring it to the top....just bring it to the top, and make it - (negative)

FOR #2 just divide 150 by 18 to find the constant.

JUST WONDERING, WHAT GRADE ARE YOU IN Im a freshie in HS

Porportions are the easiest thing I've ever done I use em' a lot. (I didn't read all ur' post, just saw the word porportion).

Lets say ya got 53 problems right out of 163 (pretty bad...). Let's figure ur' grade.

53

I'm not sure this is the best way to revise but here we go...

Television set is a square.
Let A=area of square and d=diagonal of a square;

a ) Equation connecting A and d would be : A=1/2xd2;
b ) A= 1/2x20x20; => A=200 square inches;

It's not advanced maths, it's elementary stuff really.

Thanks for the help people, finally understand some things better. I've had a bad maths teacher that makes things very hard to understand, so even this is better than a maths lesson



I'm in England, so I don't know what grade I would be in, although im in Yr. 11 (ages 15-16).



I'm aware of that, I meant that I don't know any advanced maths so I don't want extremely complicated answers

Again, thanks for the help.

I dunno if this can be of much help, but the point for the proportions problem can be solved this way:

If A is directly proportional to the square of d, then A/d^2 will be a number that will always be the same. In that case, it would be 90/15^2. Let us call that constant T-Virus.
Now, if the area A is another number, then the value of d must be a number that makes this equation valid.

A=T-Virus.
d^2

Now, why this? You know the value of T-Virus, and if A is pretty big, then d^2 must be pretty big so the division between them gives me T-Virus. And if A is small, then d^2 must be small so when you divide them you get T-Virus.

Now, back to the problem.

Replacing the elements you have it should be something like this:
A=T-Virus
20^2


Now you just solve the equation.
I'm gonna help you out in a little while, because right now I'm studying for a Calculus "mini-test" (dunno how to call it ), so as soon as I'm over with it I'l write some more.

K, now I'll help you with the rest.





Do you remember that the equation when you had a direct proportion was
A=constant?
B


Well, for the inverse proportion it is something like that, specificaly A*B=constant. Now, why this? Because, supposing we now the value of the constant, if A is very big then B must be quite small so the product of both is the constant. Obviously we can do the backwards case, if B is big.


Now, if you look at the problem you can say that:
D*P=constant. We'll call it "P.I.M.P".
So:
18*150=2700. Now, P.I.M.P=2700.

Then, the equation that connects D and P will be:
D=2700
P







When you've got something like this, the last thing you want is a square root in the lower part of the fraction, right? So, how do we get rid of it? The answer is pretty simple: you can multiply by a certain number in the upper and lower part of the fraction. For example:

1 * 3= 3
2 3 6

3 and 1 are the same right?
6 2

Now, as I said before, you gotta multiply the fraction by a number in both parts of it. And that number has got to be sqrt(6), because:
sqrt(6)*sqrt(6)=sqrt(6*6)=sqrt(36)=6.

(NOTE: Technically, sqrt(36) can be -6 or 6, because -6*-6=36 and 6*6=36, but in general terms when you work with this kind of exercises you just use the positive term, 6).

The result of this exercise would be:
sqrt(6)
6



In another note, what if the teacher asks you to rationalise
1 ?
sqrt(1-sqrt(x))

We do the same thing: multiplying in the upper and lower part of the fraction by sqrt(1-sqrt(x)) we get:

sqrt(1-sqrt(x))
1-sqrt(x)

Now, how do we get rid of that evil square root? This is the time when you gotta remember this thing:
A^2-B^2=(A + B )*(A - B )
because:
(A + B )*(A - B )=A^2+AB-AB-B^2= A^2 -B^2.

With that case it's just the same thing, with A=1 and B=sqrt(x). So, instead of multiplying for a square root or something like that, you multiply in both parts of the fraction for 1+sqrt(x), and you get the following:

sqrt(1-sqrt(x)) * (1+sqrt(x))
1-x

Problem solved!



I hope this has been of any help. If you've got any more questions, just ask

Ah, I was so crap at maths. I am glad that my maths GCSE is far enough behind me for me not to sweat when I think about it.