Foundation Maths Paper 2.

fufureida

Ok so I was looking through 2007 and 2008 and I can hardly attempt 4 questions let alone 6... I really need help if someone could write up a list of definite questions I could practise today? I just want to pass the paper... What is a must know and which questions do you guys reckon I should practises? Please help I'm a repeat and failed maths last time I don't wanna fail it again and basically just want to pass...:(

fufureida

Please anyone...???

Appleglue

Man, Download and print paper 2 off

www.examinations.ie

answer the questions, if your stuck then download the marking scheme as well, then just look at that and it will show you the answer and method.

in the exam, you'll have all the formulas!

its tough to recommend what questions to do, cus I find a lot of them easy. I'm doing foundation because I'm lazy, and I wanted to focus on other subjects

2,3,4,5,6,7 seem pretty basic in paper 2 2007, just use the formulas when needed and take your time tomorrow

good luck!

fufureida

Appleglue wrote: »

Man, Download and print paper 2 off

www.examinations.ie

answer the questions, if your stuck then download the marking scheme as well, then just look at that and it will show you the answer and method.

in the exam, you'll have all the formulas!

its tough to recommend what questions to do, cus I find a lot of them easy. I'm doing foundation because I'm lazy, and I wanted to focus on other subjects

2,3,4,5,6,7 seem pretty basic in paper 2 2007, just use the formulas when needed and take your time tomorrow

good luck!

Oh I went and did a paper and get stuck on stupid little questions... Currently stuck lol. Maths is not my ball game...:(

FridaysWell

fufureida wrote: »

Oh I went and did a paper and get stuck on stupid little questions... Currently stuck lol. Maths is not my ball game...:(

I don't do foundation I do ordinary, but I think I can help.

What are you stuck on? What areas are you weak/strong at?

fufureida

I'm weak at everything lol. I'm currently stuck on q4 part c 2007

I managed to do everything so far

fufureida

Ok I have managed to do a bit and stuck again... Lol

FridaysWell

fufureida wrote: »

I'm weak at everything lol. I'm currently stuck on q4 part c 2007

I managed to do everything so far

(c) The line K has equation y = 3x + 2 .
The line L has equation 2y = 6x −1.
The point c has co-ordinates (1, 5).

(i) Show that the point c lies on the line K.
For this, you substitute (1, 5) into the equation of the line.
You do this because if the point lies on the line, it will statisfy the equation.
That means you replace y with either 1 or 5 and you do likewise with x.
So, here goes:
y = 3x + 2
5 = 3(1) + 2
5 = 3 + 2
5 = 5
THEREFORE THE POINT (1,5) is on the line K.

(ii) Write down the slope of K.
If you know the equation of a line, which is in the log tables it will say: y - y1 = m(x -x1)
m = slope
K = y = 3x + 2, as you can see, K is in the same format.
So from K we can see the slope is 3

(iii) Write down the slope of L.
Applying the same thinking and principle of part (ii) we can see the slope for L is 2y = 6x - 1
To symplify this we must divide by 2, because the y value must be y only, with no number.
So, we get, y = 3 - 1/2
Therefore the slope for L is 3

(iv) Are the lines K and L parallel? Explain your answer
If lines are parallel, the slope of the two lines will be the same.
Slope K=3
Slope L=3
Therefore they are parallel


:D

fufureida

Omg I gotta go over this for awhile... Thanks so much though!!

FridaysWell

fufureida wrote: »

Omg I gotta go over this for awhile... Thanks so much though!!

Grand, any Q's just ask and I'll explain

fufureida

I thought simultaneous equations worked to see if c was on the line?

Also I don't get how you got tge slope... How do I apply rye actual slope formula to it???

FridaysWell

fufureida wrote: »

I thought simultaneous equations worked to see if c was on the line?

Also I don't get how you got tge slope... How do I apply rye actual slope formula to it???

Using simultaneous equations would really complicate a simpl question.
All you need to do is sub C into the line equation, and if one side equals the other, it is on the line.

Thats the beauty of the question, you don't need to apply the slope formula, the answer is there, in front of you.

fufureida

FridaysWell wrote: »

Using simultaneous equations would really complicate a simpl question.
All you need to do is sub C into the line equation, and if one side equals the other, it is on the line.

Thats the beauty of the question, you don't need to apply the slope formula, the answer is there, in front of you.

Yeah but how do I know that's the slope? like I know you are looking at it in the format at the equation of a line but I still don't get it. Sorry I'm thick at maths.

Can you help me with q5? I think I'm right for the part A just need to make sure and I'm really clueless for the part B...

FridaysWell

fufureida wrote: »

Yeah but how do I know that's the slope? like I know you are looking at it in the format at the equation of a line but I still don't get it. Sorry I'm thick at maths.

Can you help me with q5? I think I'm right for the part A just need to make sure and I'm really clueless for the part B...

If it is in the same format as the log tables then the answer is in dfront of you usually.
Its ok its not an easy subject. Just approach it with a little confidence and logic and you'll be fine.

I really would but I really have to go now.
I'm sorry, although I hoped I helped.
Best of luck tomorrow hope it goes well.

fufureida

Ok thanks!!

Carsinian Thau

5. (a) The diagram shows a right-angled triangle with sides of length 5, 12 and 13 and an angle X. (i) Write down the length of the hypotenuse. (ii) Write down cos X as a fraction.

(i) Length of hypotenuse = 13. (Side opposite the right angle).
(ii)cos X = 5/13 (The cos of an angle is the length of the adjacent side over the length of the hypotenuse).

(b) In the diagram, find (i) tan A (ii) the measure of the angle A, correct to the nearest degree.

(i) tan A = 9/4 (The tan of an angle is the length of the opposite side over the length of the adjacent side).
(ii) tan A = 9/4.
Therefore the inverse tan A (tan^-1) will give the angle. So tan^-1 (9/4) = 66.04 degrees. So correct to the nearest degree it's 66 degrees.

(c) A bird takes flight from a point p on the ground. The bird rises at an angle of 25° to the ground. After flying 40 metres, what is the bird’s height above the ground? Give your answer to the nearest metre.

sin 25 = h/40 (Let h be the height above the ground).

Therefore the h is equal to 40 multiplied by sin 25.

Or h = 40 sin 24
h = 40 * 0.4226
Therefore, h = 16.9 metres.
Corrected to nearest metre, that's 17 m.

fufureida

Omg thank you so much I got it all accept part c will look at that in the morning thank you sooo much!!!

Carsinian Thau

No problem at all. Happy to help.

Do you want a more detailed thing for part c? (Or anything else for that matter?)

fufureida

Carsinian Thau wrote: »

No problem at all. Happy to help.

Do you want a more detailed thing for part c? (Or anything else for that matter?)

More deataiiled for part c

I have not studied maths all year and that damn part c comes up every year so I really just want to be able to do it...:(

fufureida

God I feeL so screwed for maths. I don't get anything in question 1 but simosons rule like i never know the part a or c... It always confuses the crap outta me... I can hardly do 4 questions...:(

Carsinian Thau

Look at the diagram they gave us for this too:

Seeing as we have been asked to calculate the bird's vertical height above the ground we can construct a right angled triangle. We are told that the bird flies off at an angle of 25 degrees and that (s)he flies 40 metres from the starting point. This length is the hypotenuse as it is opposite the right angle (90 degrees).

From this we can work out the height as the sine of the angle they give us (25 degrees) will be equal to the length of the height over the distance it flew. This is because the length of the height is the side opposite the angle and the distance it flew is the hypotenuse.

Therefore, sin 25 = Height / 40

Multiply both sides by 40 to remove the fraction:

40 * sin 25 = Height

The sine of 25 degrees = 0.422618262 (Make sure your calculator is set to degrees!!!)

So, Height = 40 * 0.422618262

Therefore, Height = 16.9047305 metres.

Asked for answer correct to nearest metre means (in this case) rounding up to give an answer of 17 metres.

Carsinian Thau

fufureida wrote: »

God I feeL so screwed for maths. I don't get anything in question 1 but simosons rule like i never know the part a or c... It always confuses the crap outta me... I can hardly do 4 questions...:(

You really will be fine. The examiners are there to try and get you to pass. Just do your best and it really will be ok in the end.

Unfortunately, I can't help you with the Simpson's Rule thing, I did HL and it didn't seem to be on our course.

fufureida

Carsinian Thau wrote: »

You really will be fine. The examiners are there to try and get you to pass. Just do your best and it really will be ok in the end.

Unfortunately, I can't help you with the Simpson's Rule thing, I did HL and it didn't seem to be on our course.

I really hope so I just want to pass if I fail maths again my parents will never speak to me again... I'm so nervous right now I can't sleep.