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anonymous
 5 years ago
I've got a math exam on friday. I have some of the previos exams here and there are a whole bunch of questions which I cant solve. I really needs some help with them. I'll be posting them here now. Thank you very much for all who help me out, I really appreciate it :)
anonymous
 5 years ago
I've got a math exam on friday. I have some of the previos exams here and there are a whole bunch of questions which I cant solve. I really needs some help with them. I'll be posting them here now. Thank you very much for all who help me out, I really appreciate it :)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Limits problems, find the limit: a) \[\lim_{x \rightarrow 5}x/(x5)\] b) \[\lim_{x \rightarrow \infty}(\sqrt{x+1}\sqrt{x})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a) add and subtract 5 frm d numerator that gives u 1 + 1/(x5) when u put in infinity it is 1 + 1/inf = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as for b) multiply and divide by \[\sqrt{x+1}+\sqrt{x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that gives u 1 / √x+1+√x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when u put in infinity it gives u 1/inf = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a) \lim_{x \rightarrow 5} { x \over x5} =\infty\] That's because when you substitute 5 in the top you get 5, and in the bottom you get 0. A number over 0 is infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Next up, find the derivative: a) \[f(x) =x^{2}\times e^{x ^{2}}\] b) \[f(x,y)=(y ^{3}x+e ^{2y})^{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a) dy/dx= 2xe^x3 + 3x^4e^x^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b) him1618 already got the right answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the way, in a) if the limit approaches 5 from the left, it would be infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so ur saying it doesnt exist?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah neway its a sort of hyperbola centred at 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It has a vertical asymptote at x=5.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Question 3: Define the min/max of \[f(x)=x^{3}12x ^{2}+21x+25\] in [1, infinity[

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0differentiate it f'(x) = 3x^2  24x + 21

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Question 4: Define the local min/max of \[f(x,y)=x ^{3}y ^{3}+3xy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sick how good you are :) Two more questions...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Question 5: Solve the following inequalities a) \[x^{3}+2x ^{2}+2x \le0\] b) \[\log_{2} x <2+\log_{4} x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Question 6: Solve the following equations a) \[\left x3 \right+\left x+2 \right=10\] b) \[2^{3x}=3^{2x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe it's right since it's factored to x(x^2+2x+2). But the part x^2+2x+2 is always >0. So, the sign just depends on x. Hence the whole expression is <=0 when x<=0.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0anwar is like totatally right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so am i man..dont demean me..lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why is the part x^2+2x+2 always >0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because its determinant is negative...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. You're the one who got the answer. I was just following your foot steps :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is an upward parabola where the min value is greater than 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You deserve a medal him1618.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I want to solve part b) in inequalities.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Please do I can't solve it :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its damn tedious to write these eqns

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on d comp only..on paper it took a minute

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i think i considered all the cases for 6a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\log_2x\log_4x<2 \implies \log_2x{\log_2x \over \log_24}<2 \implies \log_2x{1 \over 2}\log_2x<2\] \[{1 \over 2}\log_2 x<2 \implies \log_2{\sqrt x}<2 \implies \sqrt x<4 \implies x<16\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But x has to be a positive number, So 0<x<16

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nice answer myininaya helpd me a bunch :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah 6a is bloody tedious..nice wrk

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right on the spot AnwarA I'll jot that down to my notebook :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did anyone solve 6a, or I can take it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes its been solved by myininaya

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OMG.. I don't hate you at all :P

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you can look over my work. there might be a mistake who knows

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0it looks good to me though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can hardly read it :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, It's right. I got the exact same answers.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you can also graph it in your calculator if you want to verify your answers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x=15/2 for x>=3, x=5/2 for x<=2. No solution for 2<x<3.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i can write neat when i'm not trying to beat you guys to answer the questions lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i wish in high school there was a math quiz bowl and i participated that would have been so fun

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol we should divide the questions equally. I wish I had a scanner .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One question, what happens here: \[\log_2x{\log_2x \over \log_24}<2 \implies \log_2x{1 \over 2}\log_2x<2\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0log4 to base to is 2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mm, yes of course! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you still have any questions, or that's it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think that is about it :) So quick and dirty. Thanks a million for all of you!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0yes all your questions were being answered so quickly by him hes such a meanie

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont sweat..i hate studying so much...bt i lov doing this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think that is about it :) So quick and dirty. Thanks a million for all of you!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good luck in your exam on Friday :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah good luck...remember me if u feel nervy..lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks, I believe I can make it :D
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