anonymous
  • anonymous
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 :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Limits problems, find the limit: a) \[\lim_{x \rightarrow 5}x/(x-5)\] b) \[\lim_{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})\]
anonymous
  • anonymous
a) add and subtract 5 frm d numerator that gives u 1 + 1/(x-5) when u put in infinity it is 1 + 1/inf = 1
anonymous
  • anonymous
as for b) multiply and divide by \[\sqrt{x+1}+\sqrt{x}\]

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anonymous
  • anonymous
that gives u 1 / √x+1+√x
anonymous
  • anonymous
when u put in infinity it gives u 1/inf = 0
anonymous
  • anonymous
\[a) \lim_{x \rightarrow 5} { x \over x-5} =\infty\] 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
  • anonymous
Next up, find the derivative: a) \[f(x) =x^{2}\times e^{x ^{2}}\] b) \[f(x,y)=(y ^{3}x+e ^{2y})^{3}\]
anonymous
  • anonymous
a) dy/dx= 2xe^x3 + 3x^4e^x^3
anonymous
  • anonymous
b) him1618 already got the right answer.
anonymous
  • anonymous
was a wrong?
anonymous
  • anonymous
I guess :)
anonymous
  • anonymous
By the way, in a) if the limit approaches 5 from the left, it would be -infinity.
anonymous
  • anonymous
so ur saying it doesnt exist?
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
yeah neway its a sort of hyperbola centred at 5
anonymous
  • anonymous
It has a vertical asymptote at x=5.
anonymous
  • anonymous
Question 3: Define the min/max of \[f(x)=x^{3}-12x ^{2}+21x+25\] in [-1, infinity[
anonymous
  • anonymous
differentiate it f'(x) = 3x^2 - 24x + 21
anonymous
  • anonymous
the roots are 7, 1
anonymous
  • anonymous
Question 4: Define the local min/max of \[f(x,y)=x ^{3}-y ^{3}+3xy\]
anonymous
  • anonymous
Sick how good you are :) Two more questions...
anonymous
  • anonymous
yeah go on
anonymous
  • anonymous
Question 5: Solve the following inequalities a) \[x^{3}+2x ^{2}+2x \le0\] b) \[\log_{2} x <2+\log_{4} x\]
anonymous
  • anonymous
Question 6: Solve the following equations a) \[\left| x-3 \right|+\left| x+2 \right|=10\] b) \[2^{3x}=3^{2x}\]
anonymous
  • anonymous
a) x<=0
anonymous
  • anonymous
is it right
anonymous
  • anonymous
I 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
  • myininaya
anwar is like totatally right!
anonymous
  • anonymous
so am i man..dont demean me..lol
anonymous
  • anonymous
Why is the part x^2+2x+2 always >0?
anonymous
  • anonymous
because its determinant is negative...
anonymous
  • anonymous
Yeah. You're the one who got the answer. I was just following your foot steps :)
anonymous
  • anonymous
it is an upward parabola where the min value is greater than 0
anonymous
  • anonymous
Ah I understand :)
anonymous
  • anonymous
You deserve a medal him1618.
anonymous
  • anonymous
thnx a lot
anonymous
  • anonymous
I gave one too :)
anonymous
  • anonymous
You're welcome :)
anonymous
  • anonymous
so kind
anonymous
  • anonymous
I want to solve part b) in inequalities.
anonymous
  • anonymous
b) x is in 0 to 16
anonymous
  • anonymous
Please do I can't solve it :/
anonymous
  • anonymous
thats d ans
anonymous
  • anonymous
its damn tedious to write these eqns
anonymous
  • anonymous
Yeah, that sucks....
anonymous
  • anonymous
on d comp only..on paper it took a minute
myininaya
  • myininaya
i think i considered all the cases for 6a
1 Attachment
anonymous
  • anonymous
2 is \[\log_{4}16 \]
anonymous
  • anonymous
\[\log_2x-\log_4x<2 \implies \log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2\] \[{1 \over 2}\log_2 x<2 \implies \log_2{\sqrt x}<2 \implies \sqrt x<4 \implies x<16\]
anonymous
  • anonymous
But x has to be a positive number, So 0
anonymous
  • anonymous
right
anonymous
  • anonymous
Nice answer myininaya helpd me a bunch :)
myininaya
  • myininaya
here is 6b
1 Attachment
anonymous
  • anonymous
yeah 6a is bloody tedious..nice wrk
anonymous
  • anonymous
b is right 2
anonymous
  • anonymous
Right on the spot AnwarA I'll jot that down to my notebook :)
anonymous
  • anonymous
Did anyone solve 6a, or I can take it?
anonymous
  • anonymous
yes its been solved by myininaya
anonymous
  • anonymous
I hate you man >.<
myininaya
  • myininaya
woman* lol
anonymous
  • anonymous
OMG.. I don't hate you at all :P
myininaya
  • myininaya
lol
myininaya
  • myininaya
you can look over my work. there might be a mistake who knows
myininaya
  • myininaya
it looks good to me though
anonymous
  • anonymous
I can hardly read it :P
myininaya
  • myininaya
:(
myininaya
  • myininaya
i can rewrite it
anonymous
  • anonymous
No, It's right. I got the exact same answers.
myininaya
  • myininaya
you can also graph it in your calculator if you want to verify your answers
anonymous
  • anonymous
x=15/2 for x>=3, x=-5/2 for x<=2. No solution for 2
myininaya
  • myininaya
right!
myininaya
  • myininaya
i can write neat when i'm not trying to beat you guys to answer the questions lol
myininaya
  • myininaya
i wish in high school there was a math quiz bowl and i participated that would have been so fun
anonymous
  • anonymous
lol we should divide the questions equally. I wish I had a scanner -.-
anonymous
  • anonymous
One question, what happens here: \[\log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2\] ?
anonymous
  • anonymous
log4 to base to is 2.
anonymous
  • anonymous
to base 2*
myininaya
  • myininaya
because 2^2=4
anonymous
  • anonymous
Mm, yes of course! :)
anonymous
  • anonymous
Do you still have any questions, or that's it?
anonymous
  • anonymous
I think that is about it :) So quick and dirty. Thanks a million for all of you!
myininaya
  • myininaya
yes all your questions were being answered so quickly by him hes such a meanie
anonymous
  • anonymous
no probs mate
anonymous
  • anonymous
yeah thats kind...
myininaya
  • myininaya
lol i'm kidding
anonymous
  • anonymous
dont sweat..i hate studying so much...bt i lov doing this
anonymous
  • anonymous
I think that is about it :) So quick and dirty. Thanks a million for all of you!
anonymous
  • anonymous
well thnx
anonymous
  • anonymous
Good luck in your exam on Friday :)
anonymous
  • anonymous
yeah good luck...remember me if u feel nervy..lol
anonymous
  • anonymous
Thanks, I believe I can make it :D

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