You're right until the fourth line up from the bottom. If (c-1) definitely has brackets round then 6=70(c-1) you need to multiply out to get 6=70c-70 then rearrange. You should get c=76/70 = 1.1 approx
Nov 7th 2009, 07:08 AM
but if i then subsitute 1.1 value for C then the answer should equal a= (3)?
Nov 7th 2009, 07:13 AM
don't sub in 1.1 (the answer is really 1.0857....) as this is the rounded up answer. if you put c=76/70 into the original equation then you will get a=3.
Nov 7th 2009, 07:49 AM
Originally Posted by decoy808
a=4b + (6/ c-1)- 15d
You were told by Willow T that your bracketing is faulty; should be:
a = 4b + 6 / (c-1) - 15d
6 / (c-1) = a - 4b + 15d
c - 1 = 6 / (a - 4b + 15d)
c = 6 / (a - 4b + 15d) + 1
Nov 7th 2009, 08:53 AM
Hi ineed to solve this exercise. Who can help me?
Let A subset of R (real numbers) A no empty. A is annotated if and only if exist M (positive rel) such that absolute value of x is less than or equal that M, for all x in A