Recent content by teffy3001

okay so then how do i find the value of b?

i think that's where my problem lies now...im not sure how to find that out.

um that two vectors are parallel if their cross product is 0?...

v=5i + bj thats where the b is..

i know if i set x equal to 2 that the denominator will equal 0 but then the point is not to have a vertical asymptote so how do i find the value of p?

the question is: u = 9i-6j and v = 5i+bj for what value of b would u-v point in the same direction as w = 7i-3j? can anyone help me get started on this problem, I am not looking for answers just a little bit of help because I'm confused.. thanks

after i factor the denominator though...where do i go from there?

its asking me to find the value of p so that the function won't have a vertical asymptote: g(x) = (3x + p) / (x^3 + 8) however, I'm not sure how to figure that out...any help?

um the course I am taking told me first multiply the numerator and denominator by sqrt(x^2+5) - 3 i might have done it wrong but then i got x^2+5-9/x-2(sqrt(x^2+5) - 3) i factored that to x^2-4/x-2(sqrt(x^2+5) - 3) and then eventually got (x-2)(x+2)/x-2(sqrt(x^2+5) - 3) canceled out x-2...

lim (sqrt(x^2+5) - 3)/(x-2) x->2 i hope that kinda makes sense... anyways, my question is that I've been doing this and i keep getting 0 in the denominator in the answer...after subsituting everything i got 4/ sqrt(9) - 3 = 4/0 so then is this equation just not possible?

haha no thanks.

so i would get ((x-h)/a)^2 - ((y-k)/b)^2 = 1 ?

i don't even know...this is so confusing. sorry to bother you with this...

thanks, so if i do all that...then that answers my first question too?

how do i solve for sec and tan? sorry but this calculus thing...its been a while for me