Mathematics: Post your doubts here!

[Silver Wing]

Can someone help me with question number 6(iii) in ON2010 P63 ASAP? -- http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_63.pdf

I tried looking for an explanation but couldn't find one. I'm sorry if this question has been asked and answered. I don't understand why the students is 5 -- or 5/12.

The reason for why its 5 for students is the question:
What is the diffenerce if there was 1 student instead of 5?
The 5 came from (5 combination 1) the number of ways by which the student sitting in front of Mrs. Lin is Chosen from the five present

Hope it helps,

 

[Albert Einstein]

Hi please help me out with this

 

[Kirabo Takirambudde]

I miss read th first sentance nd bursted out laughing .... anyhow

Hi please help me out with this

Very tricky but also very simple once you see past the bullshit they have put in the question:
The probability of having a girl is 0.5 and a boy is also 0.5, thus here are the 4 possible ways of have 1 girl.... btw G--> girls and B--> boy
(G,B) or (B,G) or (B,B,G) or (B,B,B,G)
= 0.5^2 + 0.5^3 + 0.5^3 + 0.5^4
=11/16

Good luck 2morrow dude

 

[athaan]

Thank you for the explanation, now I understand it! ^^

 

[Albert Einstein]

I miss read th first sentance nd bursted out laughing .... anyhow

Very tricky but also very simple once you see past the bullshit they have put in the question:
The probability of having a girl is 0.5 and a boy is also 0.5, thus here are the 4 possible ways of have 1 girl.... btw G--> girls and B--> boy
(G,B) or (B,G) or (B,B,G) or (B,B,B,G)
= 0.5^2 + 0.5^3 + 0.5^3 + 0.5^4
=11/16

Good luck 2morrow dude

Thnks

 

[snowbrood]

 

[snowbrood]

View attachment 17951

answer to q15 is 2.42ms^-1

 

[littlecloud11]

How do you find the vector cross product of the normal of two planes? like (1,-2,3) and (2,1,0)

 

[littlecloud11]

How do you find the vector cross product of the normal of two planes? like (1,-2,3) and (2,1,0)

Never mind. Got it.

 

[ahmadashraf]

hi guys i am new to as level i am applying 4 math as in june p1 & m1 and i need to know the best book to buy

 

[athaan]

I only use the official CIE Pure Math 1 and I'd say I done quite a good job, although I don't think I would got a full mark, ha!
I haven't done Mechanics yet, but I'm sure the official book is OK.

The most important advice from me is to work out the past papers if you want to excel in the exam.

 

[Kirabo Takirambudde]

hi guys i am new to as level i am applying 4 math as in june p1 & m1 and i need to know the best book to buy

Please what ever you do, DO NOT use the Cambridge mechanics book by quadling, any book that teaches you the topics will be good, but once you have learn them, just like athann said, u must immediately jump into past papers, especially for 'vectors' and 'work energy and power'

 

[ZainH]

Can someone answer this question?

The tangent at P to the curve y= x^2 has a gradient: 3 . Find the equation of the normal at P.

I think it has something to do with simaltaneous equations, but I'm sorta stuck :\

 

[VelaneDeBeaute]

Can someone answer this question?

The tangent at P to the curve y= x^2 has a gradient: 3 . Find the equation of the normal at P.

I think it has something to do with simaltaneous equations, but I'm sorta stuck :\

If you differentiate the equation for the curve w.r.t y, you get dy/dx = 2x.
Since this is the gradient for the curve, at P it would be equal to 3.
So the value of x at P would be 3/2. Hence put this value in the equation, and the value of y is 9/4.
The gradient for the equation of normal is indefinitely -1/3.
So the equation becomes
y - 9/4 = -1/3 ( x - 3/2)
Simplify it to get the answer!

 

[VelaneDeBeaute]

hi guys i am new to as level i am applying 4 math as in june p1 & m1 and i need to know the best book to buy

Pure Mathematics 1 by Douglas and Quadling is amazing for clearing the concepts and moving on. For Mechanics, I used both the Douglas and Quadling's Mechanics 1 and the Mechanics for A'Levels (Green book by probably the same authors)! For Mechanics especially, I'd advise to go for the past papers, because these books were crap!

 

[snowbrood]

View attachment 17951

help people please

 

[falcon678]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_S08_qp_01.pdf

question 6 part (i) please..!
thank u...!

 

[Minato112]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_01.pdf

question 6 part (i) please..!
thank u...!

f`(x) ----> Differentiate

f(x)=(3x+2)^3 - 5
f`(x) = 3(3x+2)^2 (3)
= 9(3x+2)^2

Now for all real values of x,
(3x+2)^2 > 0
9(3x+2)^2 > 0

Since the derivative of f (x) is greater than 0, f (x) is an increasing function.

 

[falcon678]

f`(x) ----> Differentiate

f(x)=(3x+2)^3 - 5
f`(x) = 3(3x+2)^2 (3)
= 9(3x+2)^2

Now for all real values of x,
(3x+2)^2 > 0
9(3x+2)^2 > 0

Since the derivative of f (x) is greater than 0, f (x) is an increasing function.

thanks....!

 

[VelaneDeBeaute]

help people please

Alright, it's a bit lengthy, so have patience with me!
First, note that A travels 3m and B travels 2.8m. This means that for the first 2.8m of the journey, A travels with exactly the same speed as B. After 2.8m, the Tension in the string eliminates, which causes A to decelerate. We have to find this speed, after the deceleration.
So, we start off by construction of two equations for A and B for the first 2.8m
For A, it is
T - 20 = 4a
For B, it is
30 - T = 3a
Add both equations, T gets eliminated, and the value of a retained is 10/7 m/s^2.
Now you find out the speed of A after 2.8m. The value of a we found may be helpful as we could simply use the formula 2as = v^2 - u^2. Remember to take s as 2.8m, and not 3m. The value of v hence comes out as 2.82 m/s.
Now, as T is no more in the string, put T = 0 N in the equation we built for A. This gives us -20 = 4a, hence the value of a now becomes -5 m/s^2
Use these values now to find the final speed. u would be 2.82 m/s, a would be -5 m/s^2, and s would be 0.2m.
The answer comes out as 2.46 m/s. There might have been a problem with me as I rounded off figures for easy calculation. If you use exact figures, you'll get exactly the same answer. For this question, I suspect that you'll have to use g as 9.81 m/s^2. However, you may know better!
I hope this helps!