Right I am practising a few last things for my Maths Exam on Monday, I've been doing loads of past papers but there is this just 1 question which I want to make sure I got correct (Its a A* question worth quite a bit of marks). So any help or input would be appreciated . Here is the question:



Sorry the drawing is bad, did a quick photoshop job to represent the image in the exam paper
Right now above is the information that is given. Here is what I have achieved so far. (Please note that this is on the Non-Calculator Paper ):

1. Label what angle we want:

2. Write in the algebra using the information given. (x represents the value of PB):

3. Since the Sine rule is Opposite over Hypotenuse we need to label them in to our triangle:

4. We need to find the value for the Opposite side. Time to do some Pythagoras:

5. So now I divide Opposite over Hypotenuse:

6. Right now we can consider that the x's cancel each other out becuase whatever number you put when you simplify the fractions you get:

Thats what I had so far. My question is that is the answer k = 4? And if so how would you explain the last step becuase you can't exactly say that they cancel each other out. Or can you?

Thanks
- Suhail

Yup, looks right. If you solve for your angle B (using inverse sin), you get 53 degrees, which is reasonable.

You can cancel the x's in the last step.

Good luck Monday.

sin=opp/hyp
cos=adj/hyp
tan=opp/adj

Using the inverse sin function, where B=4/5, the angle comes out to be 53.13º.

I'll explain the rest:

Hope that helps.

Good job Numberzz. A way to remember the sin/cos/tan stuff is this: Soh Cah Toa (Pronounced So Ka Toe-Ha)
The letters stand for:
S > Sine
O > Oppisiter
H > Hypotenuse
A > Adjacent

No, that's a dumb way to remember. The correct way is:

I also remember SOHCAHTOA but Numberzz method seems more fun
Anyway thanks everyone. Good luck to anyone else on their exam period

PS: I don't think ya need to calculate the actual angle becuase its the non-calculator paper. Also it just says show that sin B = k/5 but thanks anyways

4x / 5x is the same as: 4 / 5 * x / x . Therefore you can cancel out the x's as any number over itself is equal to 1. So 4 / 5 * 1 = 4 / 5. That means you are correct in saying the x's cancel out.


I've graduated in Calculus so any algebra, differentiation, integration, trig etc. questions feel free to ask.

Massive thanks to everyone who helped on the previous question. Trust me you guys and gals rock .
Right another question. This one is to do with Quadratics and Fractions + Equations on the Calculator Paper

The Question:


So we need to find the 2 values of x

1. Cross Multiply the fractions. That becomes:
2. Now simplify out the brackets:
3. Now multiply the left section to remove the fraction and simplify:
4. Now we want a quadratic so everything has to be on one side to make ax^2 + bx + c = 0:
5. Now to put it in the quadratic formula:
6. The answers of x comes out to be:

Are they the right answers for x because when I put them back in to the original equation it doesn't add up correctly? Have I done something Stupid?

Thanks
- Suhail

There is something wrong:take a nice look...


So we need to find the 2 values of x

Good so far

2. Now simplify out the brackets:

(x^2+9x-12)/(2x^2-x-3)=1 ->Here is the detail Suhail (bold).


3. Now multiply the left section to remove the fraction and simplify:

x^2+9x-12=2x^2-x-3

4. Now we want a quadratic so everything has to be on one side to make ax^2 + bx +

x^2-10x+9=0


5. Now to put it in the quadratic formula:


6. The answers of x comes out to be:

x=9 or x=1

Wow thanks XxP/TuX, I did the same working out as you on my piece of paper but one mistake I made was when I moved everything from right to left. 12--3 = 9 not 15 . Which is what I did.

Also I think that + x is a typo. Its meant to be - x. Whoops (How embarrasing ). I really should learn how to use LaTeX and not Photoshop to write my equations.

Anyway massive thanks mate. . (Goes to finish off Completing the Square )
This is now solved

@Suhail No problem

@Tim Smart/All

Curiosity:there is any chance of integrate erf(x)?

I think so yes. Its been a while since integration.



Which is very intergratable. So yes is the answer

Oh,I'm sorry because I was wrong. I mean erf(z) without any limit,an undefined integral like:

Integrate(z^2,z)=z^3/3.

So it will be

Integrate(e^(-z^2),z)=?.

Is this possible?