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Mathematics: Post your doubts here!

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There are 3 ways to sit Mrs. Brown in the first row.

Once Mrs. Brown is placed in the first row, there are 10 ways in which Mrs. Lin sits behind a student. There're 5 possible students behind which Mrs. Lin sit. So the no. of ways in which Mrs. Lin sits behind a student are 10 * 5.

Now there're 9 Passengers left to be seated in 11 seats. The no. of ways of seating 9 passenger in 11 seats are : 11P9.

So the total ways for seating passengers in this way are: 3* 10 * 5 * 11P9

The total no. of ways for seating passengers with no restrictions are: 14P12

The probability is : (3* 10 * 5 * 11P9) / 14P12 = 0.06868 ~ 0.0687 (3sf) Ans.
 
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Rizwan Javed Wow you're really at good at permutations and combinations. I find them rather difficult actually so any general tips that you could pass on?
The only thing to remember here is that if there's some kind of arrangement involved, then permutations must used. But if you're to select something the you're to use the combinations. You can refer to this website for learning this topic: http://www.mathsisfun.com/combinatorics/combinations-permutations.html . If you want to master this topic, solve as many questions as you can. I also used to be bad at this topic, so I solved a lot of questions on this topic and those questions which i solved were much much harder than these. So now as i had solved the harder ones, I don't feel problem in these easier ones. So just practice, that's all that I can say. Best of luck!
 
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The only thing to remember here is that if there's some kind of arrangement involved, then permutations must used. But if you're to select something the you're to use the combinations. You can refer to this website for learning this topic: http://www.mathsisfun.com/combinatorics/combinations-permutations.html . If you want to master this topic, solve as many questions as you can. I also used to be bad at this topic, so I solved a lot of questions on this topic and those questions which i solved were much much harder than these. So now as i had solved the harder ones, I don't feel problem in these easier ones. So just practice, that's all that I can say. Best of luck!
Thanks!!!

If possible then can you please post the harder one(s) over here.Once again Thanks
 
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The only thing to remember here is that if there's some kind of arrangement involved, then permutations must used. But if you're to select something the you're to use the combinations. You can refer to this website for learning this topic: http://www.mathsisfun.com/combinatorics/combinations-permutations.html . If you want to master this topic, solve as many questions as you can. I also used to be bad at this topic, so I solved a lot of questions on this topic and those questions which i solved were much much harder than these. So now as i had solved the harder ones, I don't feel problem in these easier ones. So just practice, that's all that I can say. Best of luck!
Alright thanks!
 
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Prove that the 13th term of an A.P. is zero, if five times it's fifth term is equal to 8 times its eighth term. Anyone?
Let the first term of the A.P be T1.
d is the common difference.

The 5th term would then be: T1 + d(5-1) = T1 + 4d
The 8th term would be : T1 + d(8-1) = T1 + 7d

5(T1 + 4d ) = 8(T1 + 7d)

^ Solve it and get 'd' in terms of T1.

d = (-1/12) T1

The 13th term would be:

T1 + d(13-1)

substitute the value of 'd' we found in terms on T1.

T1 -1/12T1(13-1)
= T1 - T1
= 0 (shown)
 
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Let the first term of the A.P be T1.
d is the common difference.

The 5th term would then be: T1 + d(5-1) = T1 + 4d
The 8th term would be : T1 + d(8-1) = T1 + 7d

5(T1 + 4d ) = 8(T1 + 7d)

^ Solve it and get 'd' in terms of T1.

d = (-1/12) T1

The 13th term would be:

T1 + d(13-1)

substitute the value of 'd' we found in terms on T1.

T1 -1/12T1(13-1)
= T1 - T1
= 0 (shown)

Wow. Thanks... a lot.
 
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Can anyone tell me why Tan (98.8) is equal to -6.459.. But when I do tan inverse (-6.459) then it gives me -81? Crazy..
 
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Can anyone tell me why Tan (98.8) is equal to -6.459.. But when I do tan inverse (-6.459) then it gives me -81? Crazy..
The problem is that there are infinitely many values of x such that tan(x) = -6.459
Some examples are:
x = ..., -261.2, -81.2, 98.8, 278.8, 458.8, .....

Doing tan on all the above numbers will give you answer as -6.459. So poor calculator, which one should it tell you when you perform inverse tan on -6.459…?

We have therefore defined something known as the PRINCIPAL VALUES of the inverse trigonometric functions. These are ranges from which an inverse trig function should output an answer:

sin-1(x) always gives answer between -90 and +90
cos-1(x) always gives answer between 0 and +180
tan-1(x) always gives answer between -90 and +90

So this is why your calculator gives answer as -81.2, since it falls within the range of principal values.
 
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The problem is that there are infinitely many values of x such that tan(x) = -6.459
Some examples are:
x = ..., -261.2, -81.2, 98.8, 278.8, 458.8, .....

Doing tan on all the above numbers will give you answer as -6.459. So poor calculator, which one should it tell you when you perform inverse tan on -6.459…?

We have therefore defined something known as the PRINCIPAL VALUES of the inverse trigonometric functions. These are ranges from which an inverse trig function should output an answer:

sin-1(x) always gives answer between -90 and +90
cos-1(x) always gives answer between 0 and +180
tan-1(x) always gives answer between -90 and +90

So this is why your calculator gives answer as -81.2, since it falls within the range of principal values.
One question had two answers for tan x as it was quadratic and I was only getting one correct answer while marking scheme had both. What should we do in that situation?
 
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Need help on this question ! :( I couldn't get the answer after many times of trials,

2010 O/N paper 33 question 8 ii) b)
 
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