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 :)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Limits problems, find the limit: a) \[\lim_{x \rightarrow 5}x/(x-5)\] b) \[\lim_{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})\]
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
as for b) multiply and divide by \[\sqrt{x+1}+\sqrt{x}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question