Recent content by Dank2

Answers are x=0, x=-1. Anyone have a clue how to show it ? it doesn't have to be a proof.

dont mind the t=0, since it cannot be.

You mean write down 25 as 10log1025?

How can i continue from here, answer is x=0,-2

Thanks, solved using cosine for the two triangles

AB=AC. P is on ac such that AP=3PC. Q on CB such that CQ=3BQ. Need to find the length of PQ. I know i can use the Cosine theorem, but the answer is without Cosine.

Homework Statement A has n elements. B={0,1,2,3} {1,2,3}⊆range(f) Homework EquationsThe Attempt at a Solution So in each function we must choose those 3 numbers in the range. So let's first choose all the diffrent possiblites to choose those 3: n*(n-1)*(n-2) now for the remaining elemnts, we...

I'm assuming it's true for ##n##. but by subtracting greater power of ##2k## that is less than ##n+1##, how do i get to ##n##? since ##n-####2^k## > ##1## for example if ##n =10, k=3##. ##n+1-2^3=3##.

<Moderator's note: Moved from a technical forum and thus no template.> for every natural n there exists natural k. and numbers={a0,a1,a2,...ak}∈{0,1}. so that n=i=0n∑ ai2i I will assume n=k, i know that if n is even then a0 =0. so if i assume it is true for n that is Even: n+1=i=0n+1∑ ai2i...

H'(c)= (h(b)-h(a))/b-a > 0

That there is point c where the derivative of h'(c) is parallel to the straight line connecting the two end points of the segment

if f(c)<g(c) then h(c) < 0, doesn't help me much

yes, they are differentiable. https://www.physicsforums.com/threads/proof-inx-x-1.900384/#post-5667094 i proved it in other way in the link above post 9. howcan i use mean value theorem to show it ?

that's right, so: f(x) = x-1 - Inx . f''(x) = 1 - 1/x = 0 ==> x=1. that should be either maximus or minimum f(1) = 0. now if we take the second derivative: f''(x) = 1/x^2 which means the derivative is increasing constantly and f(1) is the minimum point inthe graph, and hence x-1>=Inx

solved. thanks