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Mathematics: Post your doubts here!

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I understood every step here except one.

How did you go from

r.n = a.n
to
ax + by + cz = d

I am really sorry for disturbing but I have big trouble with r.n = a.n form of plane.

iKhaled or anyone please.
r=(x y z) and n is the direction. a is point on the plane. hope this example helps
 

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r=(x y z) and n is the direction. a is point on the plane. hope this example helps
OOOhhhhh okaaayYY!
a small question ( i am sorry it is kind of silly) what does a represent?

Thanks alottttttttttttt!! Jazaki Allah khairan....!! May Allah S.W.T reward you with highest results, and have mercy on you and your family , Aameen!!! Thank you soooo muuucccchh!!!
 
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OOOhhhhh okaaayYY!
a small question ( i am sorry it is kind of silly) what does a represent?

Thanks alottttttttttttt!! Jazaki Allah khairan....!! May Allah S.W.T reward you with highest results, and have mercy on you and your family happy, Aameen!!! Thank you soooo muuucccchh!!!
a is a point on the plane(whose equation u're finding) , for instance position vector of A that lies on the plane. Ask if u dont get. i'l do a qs for u. u help others so much, i'll b glad if i'm of sum help to u
 
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okay look
let's assume these 8 lines below are the 8 tiles
__ __ __ __ __ __ __ __

the question states that each of the tiles has an equal probability so p(black) = p(white) = p(grey) = 1/3
now the restriction is that we can't have two tiles of the same colour placed next to each other.
so let's assume that the first tile is black, hence the next one can either be grey OR white ( 1/3 + 1/3) = 2/3
since there are 7 tiles we have to place after the first one we simply do (2/3)^7
(each of the 7 remaining tiles only have an option of 2 colours, the first could be any of the three)

Hope it helped !
 
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http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s09_qp_3.pdf q7i) can some1 give me detailed procedure of how we can obtain roots of 1-3^o.5i and -1-3^05i
ok we have z^2 + (2√3)i z -4 = 0 if u notice this is a quadratic equation so we use the formula -b ± √(b^2-4ac)/2a to solve it
a= 1 b = (2√3)i and c = -4

substitute this in the equation and dont forget that i^2 = -1

if u still didn't get it and want me to solve it pls let me know
 
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http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s09_qp_3.pdf q7i) can some1 give me detailed procedure of how we can obtain roots of 1-3^o.5i and -1-3^05i

One very important note for these kind of equation where one or more of the coefficients (a, b and c in the quadratic formula) are imaginary, (in this case the coefficient of z = (2√3)i) is that the rule of conjugate pairs being roots to the equation doesn't follow.

To solve this equation, you have to just use the quadratic formula. For this solution, the values are:

a = 1
b = (2√3)i
c = -4
4ac = -16
b^2 = 2 * 2 * √3 * √3 * i * i = 4 * 3 * -1 = -12

Putting these in the quadratic equation gives us

z = (-(2√3)i +- √(-12 + 16))/2

z = (-(2√3)i +- 2)/2

z = -√3i + 1 OR -√3i - 1

Hope this helped!

Good Luck for your exams!
 
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One very important note for these kind of equation where one or more of the coefficients (a, b and c in the quadratic formula) are imaginary, (in this case the coefficient of z = (2√3)i) is that the rule of conjugate pairs being roots to the equation doesn't follow.

To solve this equation, you have to just use the quadratic formula. For this solution, the values are:

a = 1
b = (2√3)i
c = -4
4ac = -16
b^2 = 2 * 2 * √3 * √3 * i * i = 4 * 3 * -1 = -12

Putting these in the quadratic equation gives us

z = (-(2√3)i +- √(-12 + 16))/2

z = (-(2√3)i +- 2)/2

z = -√3i + 1 OR -√3i - 1

Hope this helped!

Good Luck for your exams!
thx alot .. it really really helped me
 
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okay look
let's assume these 8 lines below are the 8 tiles
__ __ __ __ __ __ __ __

the question states that each of the tiles has an equal probability so p(black) = p(white) = p(grey) = 1/3
now the restriction is that we can't have two tiles of the same colour placed next to each other.
so let's assume that the first tile is black, hence the next one can either be grey OR white ( 1/3 + 1/3) = 2/3
since there are 7 tiles we have to place after the first one we simply do (2/3)^7
(each of the 7 remaining tiles only have an option of 2 colours, the first could be any of the three)

Hope it helped !

thanks alot :)
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf
q8. anyone please explain why A.P sum formula is not used? :eek:
a=1000 and d=50
thnx in advance

Well I have no idea what A.P sum formula is... :D
But model 1 here is an arithmetic sequence, and model 2 is a geometric sequence...

For model 1 you have to find the sum of the first 40 terms of the arithmetic sequence... That's the formula..
Sn = n/2[2a1+(n-1)d]
Sn = 40/2[2(1000)+(40-1)(1000)]
Sn = 20[2000+39000]
Sn = 20[41000]
Sn = 820000

Then multiply it by 5%
5/100(820000) = 41000

Now for the geometric sequence you use the sum of 40 terms with the respective formula... Which is..
Sn = a1(1-r^n)/1-r
5% is the increase, so we'll use 110% for ratio, which is 1.1 for short.
Sn = 1000(1-1.1^40)/1-1.1
Sn = 442592

Now do the same as you did above with the arithmetic, 5%...
5/100(442592)
= 22129
3 significant figures...
= 22100
 
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They didn't ask for the unit vector, they just asked for the vector and its magnitude, I guess I forgot to mention the magnitude...
Well the magnitude is the square root of the sum of i^2, j ^2, k^2. Which is √(i^2+j^2+k^2).
Which is √(4^2 + 4^2 + 5^2) = √57.

Unit vector would be another story, where you divide the vector itself by the magnitude.
Oh alright
Thanks :)
 
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i) Since
y = asin(bx) + c ====> we can set x = 0 to get
y = c (Because asin(b * 0) = asin(0) = 0)
Therefore, checking the graph when x = 0 gives us a value of 3, so c = 3.
The maximum value of the sine and cosine functions for all angles is equal to 1. So taking the maximum point on the graph and setting sin(bx) to 1, we get
y = a(1) + c = a + c = a + 3
Since the maximum value of f(x) on the graph = 9,
a + 3 = 9 so a = 6
Now we should be able to subsitute any other values of y, a, c and x into the equation to get b:
At x = pi/2, y = 3 so:
3 = 6sin(b * pi/2) + 3
6sin(b * pi/2) = 0
sin(b * pi/2) = 0
Since the sine function is zero only for the even multiples of pi/2, we can assume that b = 2.

ii) Since we know that in this condition y = 0, we can substitute that value straight into the equation:
0 = 6sin(2x) + 3
-3 = 6sin(2x)
sin(2x) = -0.5
Using "Sine Inverse", we get
2x = -30 degrees.
This translates to an angle of 330 degrees (360 + (-30) = 330).
However, we know that the sine function for 2x is negative, and the sine function is negative in the third and the fourth quadrants, so we can find another value (aside from 330 degrees) for 2x in the third quadrant. This value can be obtained by:
180 - (-30) = 210 degrees -- adding -30 degrees to 180 would give us 150; sin(150) is positive, so 2x cannot be 150 and we have to subtract fro 180 instead.
So out of 210 degrees and 330 degrees, 210 is the smaller one and this means that x = 210/2 = 105 degrees. This is equal to 1.83 radians.

5) If the perimeter of arc R1 is equal to the length of arc R2, the perimeter of arc R1 is:
r + r + r(theta) = 2r + r(theta) = r(2 + theta)
The length of arc R2 is
r(2pi - theta)
If these two expressions are equal, we have
r(2 + theta) = r(2pi - theta)
2 + theta = 2pi - theta
2 * theta = 2(pi - 1)
Cancelling out 2 from both sides,
theta = pi - 1

ii) If the area of R1 = 30,
0.5 * r^2 * theta = 30
r^2 * theta = 60
r^2 = 60/(pi -1) = 28.016
r = 5.29 cm

Area of R2 = 0.5 * r^2 * (2pi - (pi - 1))
0.5 * 28.016 * (pi + 1) = 58.01 cm^2

Hope this helped!

Good Luck for your exams!
 
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