New servers, hooraaaay! More bandwidth, more power.
what are ALL the matrix transformations i need for IGCSE?including all reflections,rotation,translation,shear,stretch,enlargement?Please??
Let the initial population be 100 and the number of years after which the population doubles let it be n.100 * 1.10n = 200Solving for n give : 1.10n = 2Hence, log 2 / log 1.1 = 7.27 years (3 s.f.)Therefore answer is 8 years. Since after 7.27 years there will only be 199.9 people.3. The answer at the end n = 12.1 , and so 1.21^12.1 x 2= 20m exactly why is the asnwer in the book 13?
ok but for the next one , the answer in my book says 12 years why?6000 x 1.11^n = 18000 1.11^n= 3thergfore log 3/ log 1.11=10.5and anywy the anwer in the book says 12 and 6000 x 1.11^12 = 20000 +
ohhhhhhhhhh lolnow i get iti kept getting the answer 480 . now i get that we gotta go below too . THANKS !
You made a little mistake there dude. It actually is similar to a general progression The required solution :6000 x 1.11(n-1) = 3(6000)Simplify it and you'll be getting n = 11.5 which can be taken as 12. But the best answer is 11.5Same principle is applied for question 3 and 1.3) 2 x 1.21(n-1) = 20You just need to simplify and you'll get n = 131) 100 X 1.1(n-1) = 2(100) -----> n = 8.27 which can be taken as 8 Hope it helps
i never heard the thing about the n-1 but it seems to get the right answers ! thx much !!i have another questionA wallet contains $40 and has 3 times as many $1 notes as $5 notes . Find the no. of each kind.
Anytime Given the ratio of $3:$5 = 3:1 ---->$40 represents (3 + 1)=4 sharesHence 1 share represents 40/4 = $10Number of $1 = 10Number of $5 = 3(10) = 30Hope it helps