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

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- jamiebookeater

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- 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

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

as for b)
multiply and divide by \[\sqrt{x+1}+\sqrt{x}\]

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## More answers

- anonymous

that gives u 1 / √x+1+√x

- anonymous

when u put in infinity
it gives u 1/inf = 0

- 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

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

a)
dy/dx=
2xe^x3 + 3x^4e^x^3

- anonymous

b) him1618 already got the right answer.

- anonymous

was a wrong?

- anonymous

I guess :)

- anonymous

By the way, in a) if the limit approaches 5 from the left, it would be -infinity.

- anonymous

so ur saying it doesnt exist?

- anonymous

Yep.

- anonymous

yeah neway its a sort of hyperbola centred at 5

- anonymous

It has a vertical asymptote at x=5.

- anonymous

Question 3:
Define the min/max of \[f(x)=x^{3}-12x ^{2}+21x+25\] in [-1, infinity[

- anonymous

differentiate it
f'(x) = 3x^2 - 24x + 21

- anonymous

the roots are 7, 1

- anonymous

Question 4:
Define the local min/max of
\[f(x,y)=x ^{3}-y ^{3}+3xy\]

- anonymous

Sick how good you are :) Two more questions...

- anonymous

yeah go on

- anonymous

Question 5:
Solve the following inequalities
a) \[x^{3}+2x ^{2}+2x \le0\]
b) \[\log_{2} x <2+\log_{4} x\]

- anonymous

Question 6:
Solve the following equations
a) \[\left| x-3 \right|+\left| x+2 \right|=10\]
b) \[2^{3x}=3^{2x}\]

- anonymous

a) x<=0

- anonymous

is it right

- 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

anwar is like totatally right!

- anonymous

so am i man..dont demean me..lol

- anonymous

Why is the part x^2+2x+2 always >0?

- anonymous

because its determinant is negative...

- anonymous

Yeah. You're the one who got the answer. I was just following your foot steps :)

- anonymous

it is an upward parabola where the min value is greater than 0

- anonymous

Ah I understand :)

- anonymous

You deserve a medal him1618.

- anonymous

thnx a lot

- anonymous

I gave one too :)

- anonymous

You're welcome :)

- anonymous

so kind

- anonymous

I want to solve part b) in inequalities.

- anonymous

b) x is in 0 to 16

- anonymous

Please do I can't solve it :/

- anonymous

thats d ans

- anonymous

its damn tedious to write these eqns

- anonymous

Yeah, that sucks....

- anonymous

on d comp only..on paper it took a minute

- myininaya

i think i considered all the cases for 6a

##### 1 Attachment

- anonymous

2 is \[\log_{4}16 \]

- 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

But x has to be a positive number, So 0

- anonymous

right

- anonymous

Nice answer myininaya helpd me a bunch :)

- myininaya

here is 6b

##### 1 Attachment

- anonymous

yeah 6a is bloody tedious..nice wrk

- anonymous

b is right 2

- anonymous

Right on the spot AnwarA I'll jot that down to my notebook :)

- anonymous

Did anyone solve 6a, or I can take it?

- anonymous

yes its been solved
by myininaya

- anonymous

I hate you man >.<

- myininaya

woman* lol

- anonymous

OMG.. I don't hate you at all :P

- myininaya

lol

- myininaya

you can look over my work. there might be a mistake who knows

- myininaya

it looks good to me though

- anonymous

I can hardly read it :P

- myininaya

:(

- myininaya

i can rewrite it

- anonymous

No, It's right. I got the exact same answers.

- myininaya

you can also graph it in your calculator if you want to verify your answers

- anonymous

x=15/2 for x>=3, x=-5/2 for x<=2. No solution for 2

- myininaya

right!

- myininaya

i can write neat when i'm not trying to beat you guys to answer the questions lol

- myininaya

i wish in high school there was a math quiz bowl and i participated that would have been so fun

- anonymous

lol we should divide the questions equally. I wish I had a scanner -.-

- anonymous

One question, what happens here:
\[\log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2\]
?

- anonymous

log4 to base to is 2.

- anonymous

to base 2*

- myininaya

because 2^2=4

- anonymous

Mm, yes of course!
:)

- anonymous

Do you still have any questions, or that's it?

- anonymous

I think that is about it :) So quick and dirty. Thanks a million for all of you!

- myininaya

yes all your questions were being answered so quickly by him
hes such a meanie

- anonymous

no probs mate

- anonymous

yeah thats kind...

- myininaya

lol i'm kidding

- anonymous

dont sweat..i hate studying so much...bt i lov doing this

- anonymous

I think that is about it :) So quick and dirty. Thanks a million for all of you!

- anonymous

well thnx

- anonymous

Good luck in your exam on Friday :)

- anonymous

yeah good luck...remember me if u feel nervy..lol

- anonymous

Thanks, I believe I can make it :D

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