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**Al-Allo**- Replies: 5

Hi, if we have a scalene triangle with : 112,28 and 40 degrees. and that we have a side of lenght 4 between 28 and 40. Can we find the height of the triangle without using the sine law ?

Thank you!

**Al-Allo**- Replies: 27

Hi,

I was wondering, what are Transcendental numbers ? I've never heard of such type of numbers. Thank you!

**Al-Allo**- Replies: 1

Hi, there's just one thing which I'm not sure with this problem :

There are 24 pounds of nails in a sack. Can you measure out 9 pounds

of nails using only a balance with two pans?

I answered yes and here's the reason. Consider a number X of nails which weights 24 pounds. This number X has atleast a factor of 2^3 (It can be higher.) in its factors. Why ? Because we need to divide evenly the number of nails so we can have equal weight on each side of balance.

So :

x/2=x/2

x/4=x/4

x/8=x/8

x/2=12 pounds x/4=6 pounds x/8=3 pounds

So we have the following groups (after successive divisons.):

x/2+x/4+x/8+x/8=x

Clearly, x/4+x/8 nails would give 9 pounds

Now, the only thing bugging me is that my solution is founded on the assumption that all nails must have equal weights. The question isn't clear if the nails do have all the same weight or not...

What do you think ???

THank you!

Bump!

**Al-Allo**- Replies: 1

All the proportions which we have deduced from a:b::c:d may be represented generally as follows :

ma+nb:pa+qb::mc+nd: pc + qd

For the product of the extreme terms is mpac+npbc+mqad +nqbd; which, since ad = bc, becomes mpac+npbc+mqad +nqbd; also the product of the mean terms is mpac+mqbc+npad+nqbd; or, since ad = bc, it is mpac+npbc+mqad +nqbd: so that the two products are equal.

Here are some of the proportions deduced from a:b::c:d :

(a-b):a::(c-d):c

(a+b):a::(c+d):c

and some others (I'm not putting all of them.)

I don't understand the part : "All the proportions which we have deduced from a:b::c:d may be represented generally as follows :

ma+nb:pa+qb::mc+nd: pc + qd"

What does it mean ??? I'm not following for a certain reason....

Thank you!

Yeah, that would be an idea!

bobbym wrote:

That would be 9h. Think about it, if x goes to 0 how does that affect 9h. It does not.

Well, at first i thought the same thing but wasn't too sure... ok thanks

Well, I have another question.

If had something like

lim 9h

x->0

What do you do in these occasions ? Is there a definition that tells us the actions needed ? Thank you (I'm referring to the different letters.)

bobbym wrote:

Also, there is a theorem for when there is one polynomial over another. I put it into post #13.

Yes, I saw it

Ah ok, thanks for the web page.

But... how do you know that - ∞ / 5 is -∞ ???

I didn't learn differentiation...

Btw, if you want to see the problem, go her ; http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

it's example 4 at the end

I didn't learn that... yet.

bobbym wrote:

Hi;

Did you understand what happened? I can do more examples if you need them.

Nah it's okay !

**Al-Allo**- Replies: 21

Hi, basic question here:

I have

lim 4z^2+z^6

_________

1-5z^3

x->positive infinity

Now, I know that this reduces to :

4/z+z^3

________= 0+infinity

_________

0-5

1/z^3 -5

which gives infinity over -5. Now, how do I evaluate infinity over a constant ??? Is there a fact or definition that tells us how to deal with this ?

Thank you

bobbym wrote:

According to the template we now just need to solve

we get

Now we have n1,n2 and n3 so

we have found the cube root of the RHS!

Ah ok! Well, thank you for taking your time.

bobbym wrote:

Not exactly, it is a -36.

yes

bobbym wrote:

3) Now take a look at the abc term. What do you see?

That we have 36 ???

bobbym wrote:

2) Look at the coefficient in front of c^3 in my example. It is -27. Take the cube root of it. So n3 = -3.

Following

bobbym wrote:

It is easier to use this:

1) Look at the coefficient in front of a^3 in my example. It is 8. Take the cube root of it. So n1 = 2. Follow so far?

yes

bobbym wrote:

Can you extend that to a cubed form?

Well, I don't know. Let's hope that we can....

Well, I think I could show you an example of what I was talking about. I read it somewhere :

We have :

a^2+2ab+b^2

Now we know that the first term of the root will be "a"

so...

a^2+2ab+b^2(a

"a" squared cancels the other a^2

2ab+b^2(a

Now, we know that 2ab+b^2=(2a+b)b and so if we can figure out a divisor, we will also know the root. We also know the first term of the divisor because it is always the double of the firm term of the root. (Look at the square root algorithm :http://www.basic-mathematics.com/square-root-algorithm.html)

2ab+b^2(a+b (we find, by luck I guess, that we have "b" for the second term.)

____________

2a+b)2ab+b^2

2ab+b^2

______________

0

Anyway, i was searching for something like what I described.

Ah, i was trying to find a way to find the root with only the developped form.

Ok, byt is there a sort of long division method we could use ???