I did it and I got [(sin30 * sin29)/cos30]/[(cos135*cos61)/sin135]
My problem is this is for exam for university application and we can't use calculator or the sheets with trigonometric table like value of sin30 etc, I am ok with algebra since my university is focusing more on chemistry.
Hey I solved this on my way and today I showed it to my teacher and she said it isn't right way to solve it even I got the same result
I divided 3n-1/n^2+1 with n so I got 3/n and I wrote it as 1/n/3 and put that n/3=t where n=3t and I got in exponent 6t+3 and got that e^6.
Hi, I've been doing limit problems, and just got to this problem and I can't solve it. I would love some tips; you don't have to solve my problem.
Screenshot by Lightshot
Screenshot by Lightshot
The translation in binom coefficent of 4th and 10th are mathching each other.
Find the member which doesn't have x in it.
I understand all of it but the part where (n up n-3)=(n up 9) I just don't understand how they got 12 here
In solved binominal form (4x+3)^n has two members x^4 and x^3 whose binomial coefficients are equal.
I'm kinda good in solving binomial coefficient, but I never stumbled to something like this
If its not problem for you to check these last two. I got no more of these.
If you have some kind a book with tasks like this on internet I would love to print it out so I can have some fun.
1ST answer 20/3
2ND answer -20/3