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Mathematics: Post your doubts here!
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Divide every term by cost...
So:
(sint-cost)/(sint+cost) =(sint/cost-cost/cost) / (sint/cost+cost/cost) = (tant - 1) / (tant + 1)
a ^2 = ln(4+8a^2)Hey can someone help me with this question please! Thanks
e^(a^2) = 4 + 8a^2
e^(a^2) = (1+2a^2)/0.25
e^(-a^2) = 0.25/(1 + 2a^2) (taking reciprocal)
e^(-.5a^2) = 0.5/(1 + 2a^2)^0.5
0.5 = e^(-.5a^2) * sqrt(1+2a^2)
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x^3 - x + (3^.5) / 6 = 0
Can anyone solve this please ?
It is also given that this equation is equivalent to sin 3theeta = 0.75 .
Can anyone solve this please ?
It is also given that this equation is equivalent to sin 3theeta = 0.75 .
There's a standard method to find the perp distance of a point from a line.
Write the equation of the line (4-2x, -9 + x, 9 - 2x)
Subtract the co-ordinates of the point ( 4 - 2x -3, -9 + x -8, 9 -2x -5) = (1-2x, -17 + x, 4 - 2x) ---> 1
this is perpendicular to the line
(1 - 2x, -17 + x, 4 - 2x) dot (-2, 1, -2) = 0
x = 3
Put this into 1
(-5, -14, -2)
Modulus of this is the distance.
This is a common type of question and almost always involves at least one dot product being equal to zero. In this case, the direction vector of the plane is perpendicular to the direction vector of the line since the line is on the plane. So (a,b,-3).(-2,1,-2) = 0
b = 2a - 6
we need another equation so use the point on the line (4,-9,9) in the equation of the plane (again, because the line is on the plane)
4a - 9b -27 -1 = 0
solve simultaneously.
Use sin3theta = 0.75 to calculate values of theta.
theta = 0.283
theta = 0.765
Now you have to use the equation given earlier. X = 2sin(theta)/sqrt3 with these 2 values of theta to get the first 2 solutions.
Further positive values of theta won't work because if you make a table of the function, you'll see that it keeps increasing after 0.8. So the third solution is for a negative value of theta.
theta = -1.329 and put this in X = 2sin(theta)/sqrt3 to get the last answer.
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http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_33.pdf
can someone please solve q9) (i) q8) (i) just the part of finding k and q7) (ii)
can someone please solve q9) (i) q8) (i) just the part of finding k and q7) (ii)
http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_33.pdf
Can someone please help me with Q10 (iii)?
Can someone please help me with Q10 (iii)?
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thank u so much
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jazak Allah!Check this link thanks to DaniyalK. Was my doubt too
9)i) dy/dx = 2sin2x.2cos2x.cosx - sinx.sin^2(2x)
sinxsin^2(2x) = 2sin2x.2cos2x.cosx
sinx.sin2x = 4cos2x.cosx
tan2x.tanx = 4
(2tanx/1-tan^2x)*tanx = 4
6tan^2(x) = 4
tanx = sqrt(2/3)
x = 0.685
8)i) (2x^2)/(k-x^3) dx = 1/t dt
-2/3ln(k-x^3) = lnt + c
t = 1, x = 1
-2/3ln(k-1) = c
-2/3ln(k-x^3) = lnt - 2/3ln(k-1)
t = 4, x = 2
-2/3ln(k-8) = ln4 -2/3ln(k-1)
ln(k-8)^(-2/3) = ln4(k-1)^(-2/3)
((k-8)/(k-1))^(-2/3) = 4
(k-8)/(k-1) = 1/8
k = 9
7)ii) l a + ib -10i l^2 = 4 l a + ib -4i l ^2
l z l ^2 = zz*
(a + ib - 10i)(a - ib + 10i) = 4(a + ib -4i)(a - ib +4i)
(z -10i)(z* + 10i) = 4(z -4i)(z* + 4i)
zz* + 10zi -10z*i + 100 = 4zz* +16zi -16z*i + 64
3zz* + 6zi - 6zi - 36 = 0
zz* -2z*i + 2zi - 12 = 0
(a, 2, 1) . (1,2,2)/(3*sqrt(a^2 + 5)) = cos(90 - tan^-1(2))
(a + 6)^2 = 9(a^2 + 5)(cos(90 - tan^-1(2)))^2
a = 0 or 1.94
tan^(n+2) + tan^n = u^(n+2) + u^nCan someone help me solve this question please
du/dx = sec^2(x)
dx = du/sec^2(x)
dx = du/1+tan^2(x)
dx = du/1 + u^2
New limits are 0 and 1 (tan(pi/4) and tan(0))
Int(0 to 1) ((u^(n+2) + u^n/1+u^2))
((u^n(1 + u^2)/(1+u^2)))
(u^n)
integrate
(u^n+1)/n+1
use limits
(1^(n+1)/n+1)
ii)a) (1 + tan^2(x))^2 - (1 + tan^2(x))
1 + 2tan^2(x) + tan^4(x) -1 - tan^2(x)
tan^4(x) + tan^2(x)
n = 2 so definite integral is 1/(2+1)
ii)b) rewrite it as
{tan^9(x) + tan^7(x)} + {tan^5(x) + tan^3(x)} + 4{(tan^7(x) + tan^5(x))}
you can use the result 1/n+1 three times with n = 7, n = 3 and n = 5 (each pair is in the curly brackets)
definite integral is (1/7+1) + (1/3+1) + 4(1/5+1)
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http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_62/
plzzz explain q5 part 2 and 3
plzzz explain q5 part 2 and 3
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(ii) All numbers must start with either 3 or 4 to be between 3000 and 5000 and end with 4 or 8 to be even numbers.
They can be like 3__4, 3__8 or 4__8. Cannot be 4__4 as digits aren't repeated.
Total available digits are 7 so remaining for the middle two digits are 5. There are 2 spaces so 5P2. This will be thrice as there are 3 combinations for first and last digits. Hence total numbers that can be made are 5P2*3=60.
(iii) Numbers can be 1-digit, 2-digit and 3-digit. All numbers must have 5 as last digit to be a multiple of 5 as 0 is not available. So there will be 1 1-digit number. 7 2-digit numbers as first digit of these numbers can be any one of the seven available digits as digits can be repeated so they will be 35, 45 etc. There will be 49 3-digit numbers as both first and second digits can be any of the seven available digits. These will be 485, 145 etc. Total are 1+7+49=57.
ii) split into 2 cases
1. Thousand digit is 3
1st digit: 1 choice (can only be 3)
4th digit: 2 choices ( can only be 4 or 8)
2nd digit: 5 choices(any of the remaining)
3rd digit: 4 choices(any of the remaining)
2. Thousand digit is 4
1st digit: 1 choice (only 4)
4th digit 1 choice (only 8)
2nd digit: 5 choices
3rd digit: 4 choices
Total = (2*5*4) + (5*4)
iii) Split into 3 cases:
1. 1 digit number: 1 choice (5 only)
2. 2 digit number:
2nd digit: 1 choice ( 5 only)
1st digit: 7 choices
3. 3 digit number
3rd digit: 1 choice
1st and 2nd digit: 7 choices each
Total = 1 + 7 + (7*7)
It is an arithmetic sequence with a = 0.362 and d = 0.785
25 < 0.362 + (n-1)(.785)
n > 32.38
so n = 33
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I need help in Math A2 9709_w08_3
Question 8) ii and iii
Question 10) ii and iii
I know I'm asking too much, but they are the only questions i need help in for maths
I Hope to hear from you
Question 8) ii and iii
Question 10) ii and iii
I know I'm asking too much, but they are the only questions i need help in for maths
I Hope to hear from you