Recent content by Dough

weee i got it thanks for the help!

whoops i meant for the j to be on the bottom for 3.11b its like [1/(2j)][ e^(jx) - e^(-jx) ]

hi, my question is from Modern Engineering Mathematics by Glyn James pg 177 # 17a Using the exponential forms of cos(theta) and sin(theta) given in (3.11a, b), prove the following trigonometric identities: a) sin(x + y) = sin(x)cos(y) + cos(x)sin(y) and 3.11a is: cos(x) = 0.5*[ e^(jx)...

i am not sure what else y have to do to get the answer, i wrote that out as well as -1 = x^2 + 3x -5 solved and got x = 1 or -3... but what else? dy/dx = f'(x^2 + 3x - 5)(2x + 3)

I just need a nudge in the rigth direction ais don't know where to start Let y = f(x^2 + 3x - 5) find dy/dx when x = 1, given that f'(-1) = 2 Thanks!

i don't think i got it right and it was hell of a lot of work... i might give it a shot tomorrow or later i got f(x) = -8x^3 - 50x^2 - 54x + 28 thanks tho i got one so i have teh idea now its just not makign stupid mistakes! dry, we didnt even have to do number 3... i sitll try it agian...

yay, i got the first one... on the third try after the above post, i kept makign silly errors just now... i'll let you know how i do witht he the next one, thanks for the help :D

i did that but no luck i'll give it another shot, i make mistakes osmetimes so its possible i may have done somethign wrong whihc messed everythign else up...

i've written out what I've been given: f(3/2) = 0 therefore (2x-3) is a factor of f(x), Factors of f(x) = (2x-3) and (x-2) f(4) = 50 i also wrote out ax^2 + bx + c as oyu suggested but not seeing a link... :(

Well there are actually two questions i need some help with, if you could provide a full solution so i cna see the step by stpe thing it would be nice. Question 1 A quadratic function f(x) with integral coefficients has the following properties: f(3/2) = 0, (x-2) is a factor of f(x), and...

Imagine a cube where all sides have a resistor, now remove one resistor from it and replace it with an ohmmeter, the ohmmeter reads 100ohms, calculate the resistance of each resistor. By the way each resistor is the same value. How do you go about doign this question?

yes it does, i liek to think of things logically (it makes it easier to remember what forumla to use and stuff) so, the tension in the connection between the first and 2nd toboggan would be higher than the last and 2nd one because there is more weight being pulled... correct?

Three toboggans are attached to each other with ropes. A force of 145 N[forward] is pulling the first toboggan, which has a mass of 42kg. A second toboggan, with a mass of 30kg, is attached to the first second toboggan. A third toboggan with a mass of 24kg, is attached to the second toboggan...

I was just wondering what does engineering entail. What will i know how to do when i finish a university course in engineering of any sort. With a job as an engineer what would i do?, produce things based on a schematico ro somethign of the sort provided? Thanks

Great! I got it, Thanks a lot Pengwuino(cool name too) Ok before i was doing it like this 4.2m/s × t = -75m + 3.8m/s × t + 0.075t^2 because of this i kept trying to do something stupid... like 4.2m/s = -75m + 3.8m/s × t + (0.075t^2) / t and then cancel out the power to the 2 but...