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Thanks a lot Bob. Good to know i'm improving with Logarithms.
Hi all. am I allowed to do this when using the change of base formula? Thanks in advance
Hi Bob good to know i'm understanding this correctly... This came from revision exercises my lecturer created so i'll have to check with her
I don't think I was totally clear. So my problem is. It seems like that would be a very big set? And not something I would be expected to do in a exam?
Hi all. I can't think that I would get a question like this if i'm understanding this correctly
Question: There are 5 50c coins and 3 20c coins in a tin label the 50c coins f_1 to f_5 and the 20c coins t_1 to t_3. suppose you put your hand in the tin and take out 3 coins without looking. Write down the set that represents the event "I have at least one 50c coin amongst the 3 coins that I drew"
Yes I did for the next question I had to find the car's time and it worked perfectly
Bob thanks again, you really have a talent for making things which seem complicated totally understandable I never thought of handling these fractions the same way as fractions without variables.
Thanks a lot Hixy works out fine now. Bob am I aloud to multiply the
Hi all. I can't think of any other way to do this problem any ideas would be much apreciated.
A bus travels 200km at a constant speed a car traveling 10km/h faster than the bus completes the same distance in 1 hour less than the bus, how many hours does the bus take to travel the 200km?
Thanks a lot Bob and Bobbym. I have done this many times before but the way the question was written confused me a bit
Hi Bob. Differential calculus is not part of my course this year at all. So I can find x either by completing the square or
?Hi Bob. I did exactly that up to the point
Hi Bob I have the parabola and lines equations. So to find the maximum I subtracted the line's equation from the parabola's equation but from here i'm not sure.
Hi. I am a bit confused with this question the question is "Find the coordinates of M when BA is a maximum" Should I subtract the straight line from the parabola and then
I don't understand how BA has any effect on M? Doesn't M just stay where it is on the x-axis?I had some internet problems so havn't been online. Bob lol no worries and thanks a lot That answer makes a lot of sense I definitely learned something new.
Thanks a lot for all the answers. Bob when you have
as the denumerator doesn't that become 2 and notSorry I tried entering the LaTeX. Any idea where my LaTex mistake was?
Thanks
Hi Bob. Thanks a lot that was a very nice detailed explanation. The only part I don't totally understand is where you changed the base
Should I do it like this?
Should the domain be >0? Thanks again.Sorry for not being totally clear annonimnystify. The question is write the domain of f(x) and solve the equation. Zetafunc do you mean break it up like this log_2(x-4)=log_4((x-8) sorry for writing it like this, I don't know much about LaTeX yet... If this is the way to go how would the change of base step look? should it be log_2/log_2(x-8)? Thanks a lot
Hi all. I have no idea if i'm doing this the right way, If I am where do I go from here? Thanks in advance
Thanks a lot bobbym. Didn't see your post at first as it was on page 2. g(x):y=-1/2x so then x=-5 which makes more sense as p<q and q=(-4,0). I totally forgot what to do if I have e.g x^2 +5x. So x(x+5)
x=0 or x=-5
Hi. I am posting this here because it has to do with the same graph. I have to find the point P. As this is one of the points of intersection of g(x) and f(x) I tried solving the system of equations. At the end I got x^2+5x=0 and I didn't know how to proceed. One of my previous questions was find the equation of the line which is paralel to g and passes through the point R Which is (-2,-2) which I got as y=-1/2x-3. Can I solve the system of equations of that line and f(x) and then subtract 3 from the x coordinate and add 3 to the y-coordinate?
Thanks for the quick responses. Bob thanks a lot for posting that detailed explanation anyhow because I still was a bit unsure but now I know where my mistake lies. I am studying from home at the moment and have a class with a tutor once a week which makes it very difficult when I get stuck somewhere.
Congratulations Bob!!! I am still doing something wrong as my solution to the equation always is concave down even though I can see from the graph that it's concave up. Sorry I don't usually take this long to understand something but parabolas is something totally new to me.