Mathematics: Post your doubts here!

[gagandeep]

they already said leave the answer in surds..so when u do the equation x= -b ±√(b^2-4ac) / 2a u will end up with x =1/4 (-20 + √40) or 1/4(-20-√40) u can simplify it..did u get it now ?

the answer for this question is :
x= (-12+√40)/4 and x=(-12-√40)/4

 

[*MK*skydiver :D]

Can someone please help me in this :
Q1 Mechanics 2
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_52.pdf
Thanks!

 

[soumayya]

pls help i can't solve 9709_w12_qp_31 question 2) 5^(x-1)=5^(x)-5

 

[josephsai]

View attachment 22061

thx u veryyyyyyy muchh.

 

[josephsai]

r

Question number???

its a lot

 

[HubbaBubba]

The amounts of money, x dollars, that 24 people had in their pockets are summarised by Σ(x − 36) = −60 and Σ(x − 36)2 = 227.76. Find Σx and Σx^2\

Does anybody have a guide on how to solve questions like these? They are really confusing :/

 

[Alice123]

Can any1 help with a(ii) ? thanks in advance

 

[Dug]

The amounts of money, x dollars, that 24 people had in their pockets are summarised by Σ(x − 36) = −60 and Σ(x − 36)2 = 227.76. Find Σx and Σx^2\

Does anybody have a guide on how to solve questions like these? They are really confusing :/

Sum of coded values = -60
Σx = -60 + (24*36) = 804

Σ(x-36)^2 = 227.76
Expand LHS:
Σ(x^2 - 72x + 1296) = 227.76
Σx^2 - 72Σx + Σ1296 = 227.76
We have Σx = 804 and Σ1296 = (1296)(24) = 31104
Σx^2 - (804)(72) + 31104 = 227.76
Σx^2 = 27011.76

 

[littlecloud11]

Can any1 help with a(ii) ? thanks in advance

5ai) For three courses a dish has to be picked from each course.
For the first course there are 3 options so 3C1
For the second course there are 5 options so 5C1. But each dish has two possible choices of being served with either fries or potatoes. So the actual combination is 2* 5C1
For the third course 3C1
Total no of ways = 3C1 * 5C1* 2 *3C1 = 90

ii) Only two courses can be picked. Possible choices - Starter-main, Starter-dessert, Main-dessert
Starter-main= 3C1* 5C1 *2 = 30
Starter-dessert= 3C1*3C1 = 9
Main-dessert= 5C1* 3C1 *2 =30
Total = 30 + 30 +9 =69

b) You have to find the different ways to divide not different arrangements. So, 14C5* 9C5 * 4C4 =252252.
The answer will be the same no matter which table you decide to chose first.

 

[Alice123]

5ai) For three courses a dish has to be picked from each course.
For the first course there are 3 options so 3C1
For the second course there are 5 options so 5C1. But each dish has two possible choices of being served with either fries or potatoes. So the actual combination is 2* 5C1
For the third course 3C1
Total no of ways = 3C1 * 5C1* 2 *3C1 = 90

ii) Only two courses can be picked. Possible choices - Starter-main, Starter-dessert, Main-dessert
Starter-main= 3C1* 5C1 *2 = 30
Starter-dessert= 3C1*3C1 = 9
Main-dessert= 5C1* 3C1 *2 =30
Total = 30 + 30 +9 =69

b) You have to find the different ways to divide not different arrangements. So, 14C5* 9C5 * 4C4 =252252.
The answer will be the same no matter which table you decide to chose first.

thanks a lot

 

[HubbaBubba]

Sum of coded values = -60
Σx = -60 + (24*36) = 804

Σ(x-36)^2 = 227.76
Expand LHS:
Σ(x^2 - 72x + 1296) = 227.76
Σx^2 - 72Σx + Σ1296 = 227.76
We have Σx = 804 and Σ1296 = (1296)(24) = 31104
Σx^2 - (804)(72) + 31104 = 227.76
Σx^2 = 27011.76

I understood the Σx part, but not the Σx^2 part :/ Why did you do that?

 

[Dug]

I understood the Σx part, but not the Σx^2 part :/ Why did you do that?

I expanded Σ(x-36)^2 it to pull out the term Σx^2. That is what we are supposed to find.

Consider this example:

(a+b)^2 = 4
a^2 + 2ab + b^2 = 4

If we are given 'ab' and 'b^2', we can easily find a^2 by putting the values into the equation. My solution is exactly the same. Did you get it now?

 

[Kumkum]

AOA wr wb
can someone pls help me with question 5(i)
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s04_qp_3.pdf

 

[Busy Buddy]

Assalamualikum
Can someone please help me with question 7(ii), 8(i) & 11(ii).
Thanks a bunch in advance.

 

[messi10]

Paper 3 Help needed
Paper: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf

Question No. 10 - Part (iii)

Someone please help me with it. Please explain with basic concepts.. Paper 3 is killing me!! Please please help :'(

 

[HubbaBubba]

T

I expanded Σ(x-36)^2 it to pull out the term Σx^2. That is what we are supposed to find.

Consider this example:

(a+b)^2 = 4
a^2 + 2ab + b^2 = 4

If we are given 'ab' and 'b^2', we can easily find a^2 by putting the values into the equation. My solution is exactly the same. Did you get it now?

Yes I understand now! Thank you so much!

 

[Alice123]

AOA wr wb
can someone pls help me with question 5(i)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s04_qp_3.pdf

cos^2θ=(1+cos2θ)/2
sin^2θ=(1-cos2θ)/2
cos^2(2θ)={1+cos2(2θ)}/2=(1+c0s4θ)/2
LHS=sin^2θcos^2θ
={(1-cos2θ)/2}{(1+cos2θ)/2}
=(1-cos^2(2θ))/4
={1-(cos2θ)^2}/4
={1-(1+c0s4θ)/2}/4
=(1-4cosθ)/8=RHS
Hope this helps . Ask if u dnt understand

 

[littlecloud11]

Paper 3 Help needed
Paper: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf

Question No. 10 - Part (iii)

Someone please help me with it. Please explain with basic concepts.. Paper 3 is killing me!! Please please help :'(

10 iii) You have the equation of two lines that the plane contains.
l= <1,1,1> + s<1,-1,2>
m= <4, 6, 1> +t <2, 2, 1>
The normal vector n, of the plane is perpendicular to the directional vectors of both of the lines contained in the plane. Therefore to find the normal vector you take the cross product of the direction vectors of the two lines.
n = <1,-1, 2>x<2, 2,1>
1 -1 2
2 2 1
=(-1 -4)i - (1-4)j + (2+2)k
= -5i + 3j +4k
So the normal of the plane is -5i + 3j +4k
Use r.n =a.n to find the equation of the plane. You can use the position vector of any point of the plane as 'a'. like when s= 0 the position vector of a point on line l = <1 1 1>. You can take and value of s or t to find a point.
So, <x y z>.<-5 3 4> = <1 1 1>.<-5 3 4>
-5x + 3y + 4z = 2
That's the plane equation.

 

[Jspake]

5iii) y= a- bCos x, The max and min value of Cos x are 1 and -1 respectively.
For this equation, y has a maximum value when Cos x =-1 (and b is negative) i.e. y= a -(b* -1) = a+b.
y has a min value when Cox x =1 (and b is positive) i.e. y = a- (b*1) = a-b
After you have found the max and min values you have to find the values of x at which these occurs. Cos x is equal to -1 only when x is equal to 180 in the given range so the max occurs at 180 degree. Cos x is +1 for two values of x 0 and 360, therefore the min occurs at these values.
Now sketch the curve.
View attachment 21794

Hope this helps. If you have any questions feel free to ask.

Sorry for late reply..

Thanks a lot for your help

 

[Light Yagami]

The detailed solution of:-