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Yeh I needed a general solution.
But Its ok now, went to a help lecture and got the info I needed.
Thanks for the input though
(on my sig aswell )
ah no mate, Im just after getting the origonal expression: ax=b mod m
into the form:
x = c mod m'
where apparently m' = m/hcf(a,m)
Lol, I havent logged onto this forum in a few years, I didnt reaslise I still had that siggy
Hi.
Having real problems understading my lecutre notes about this.
Can anyone do this question and explain how you get the answer please?
Thanks
Determine whether or not there is a solution; if there is, find the General Solution.
18x = 21 mod 25
EDIT -(Oh and if that doesnt have a solution heres another:
6x = 5 mod 13)
If you wish to find the indicated angle in this diagram, it is arccos(20/30). The hypotenuse is 30 because the problem tells us that the bowl is a hemisphere, so the distance from the center to any point on the surface of the bowl is simply the radius, 30. Hopefully this is the mentioned angle you needed to solve the problem, if you need more pieces of the puzzle, just ask.
Thats it mate!!
Thanks v much!
@John:
Nope the links working John, mabey its just need to load? Ill put another at the bottom of this post though.
@Ricky:
Its really a question about circular motion so you just forget about whats powering the ball. Just say its magnetism or something. And yes, gravity is involved.
The missing link im looking for is the angle which the Reaction force makes with the horizontal. With that im sorted.
I set off resolving forces around the reaction force and equating them to the weight and so on.
Gets horrible........
How do you differentiate:
1 / ln x ????????????????????????????????????????????
Sorry mate, while I was waiting for a reply I worked it out. And its just 9.8. Thanks though
darn I hate this question! Theres no diagram but Ill copy the question out. Oh and the topics Parbolic motion.
A particle is projected at a speed of 45ms at an angle of 30 degrees above the horizontal.
(a) Calculate
(i) the speed and direction of the motion of the particle after 4 seconds,
(ii) the maximum height above the point of projection reached by the particle.
I can do (b) but the first two are the ones that got me. Anyhoo the answers are:
(a)
(i) 42.4 ms at 23.2 degrees below the horizontal
(ii) 25.8 m
Why does this forum sometimes feel like a general chat zone instead of obvious topic?
hello, Just found this forum, looks very useful for my next year of painstaking futher maths.
I do Physics and IT aswell at QEHS in England.
Maths IS fun, when taught with style.......
(also you might want to know i did German for GSCE but got a "D")
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