All Answers 358115 Answers

Thanks, reinout......I think we "c" it now.......(LOL!!)......points and thumbs!!!

remember that

3c = c + c + c

7c = c + c + c + c + c + c + c

5c = c + c + c + c + c

So we can say

3c+7c+5c = c + c + c + c + c + c + c + c + c + c + c + c + c + c + c = (3+7+5)c = 15c

I haven't seen you menion that forum in a while Melody .

I see that you really want to get to the bottom of this .

ur welcome CPhill!

Thanks, rosala.......so far....you have sent me flowers and money this week....can't beat that!!

I haven't drawn it up but I think that the common area (that is the overlap) might  be

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{15}}}{{\mathtt{20}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{15}}}{{\mathtt{20}}}}\right)}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.5}}{\mathtt{\,\times\,}}{{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}\right)} = {\mathtt{166.041\: \!867\: \!567\: \!676\: \!442}}$$ 

I do not have time to check that it is right. I wish I did.  Maybe later.

CPhill uve done such good to this forum so ill pay u !lol!

a whole lot full of dollars for u !Enjoy!

thank u CPhill!

melody i havent seen u making such posts but zegroes should think about this !he gets a too friendly sometimes but let it be , im sure hes not gonna repeat it !

Yes Rosala, you are right.  Zegroes and I need to think about this before we make public posts.

Thanks, heureka, for showing how to find that oblique asymptote....I was going to go through it but I was working on another problem......besides......I couldn't have presented it better, anyway!!!

I want to point out one more thing.....the remainder of (24)/(x-3) means that we actually don't have a perfectly linear asymptote when x is relatively "small."  However, as heureka shows by limit analysis, this "fiddly" asymptotic behavior disappears as x aproaches "large" positive and negative values. Thus, we can ignore the remainder and write the aymptote as heureka has done !!!!

thank u melody !melody i too knew that zegroes was just joking!i was just trying to tell him that its not a good thing to post like that bout melody becoz just leave me , those who r new in here will have ur bad impression on them ! i know how everyone jokes with each other so i wont mind it much but there will be someoneelse who can mind it so that was the thing i tried !btw thanks !and thumbs up from me!

y = (x^2+6x-3)/(x-3)   'slant' asumtotes ?

$$y=\dfrac{x^{\textcolor[rgb]{1,0,0}{2}}+6x-3}{x^{\textcolor[rgb]{1,0,0}{1}}-3}$$

degree top = 2

degree botton = 1

'slant' asumtotes  when degree top = degree botton + 1

'slant' asumtote yes ! ( 2 = 1 + 1 )

$$\begin{array}{cccccc}
(x^2&+&6x&-&3)& : (x-3)= \textcolor[rgb]{1,0,0}{x+9}+\dfrac{24}{x-3}\\
-(x^2&-&3x)\\
0&+&9x&-&3 \\
&-&(9x&-&27)\\
&&0&+&24\\
\end{array}$$

$$\lim\limits_{ x \to \pm\infty }(\frac{24}{x-3})=0$$

'slant' asumtote  y = x + 9

Using the information in the first part of the problem, let d be the distance from C to B and h be the height of the building. So we have,

tan(25) = h/d   Solving for h we have, h = d*tan(25)

And, since we have a right triangle with two legs d  and 160, the distnce that D is from B = √(160² + d²)

So we have

tan(20) = (h)/√(160² + d²) = (d*tan(25)) / √(160² + d²)

Simpifying, we have

tan(20)/ tan(25) = d/√(160² + d²)     and we can write

tan(25)/tan(20) = √(160² + d²) /d       Square both sides

[tan(25)/tan(20)]² = (160² + d²)/d²    and the right side = 160/d² + 1   Subtract 1 from both sides

([tan(25)/tan(20)]² - 1)  = 160²/d²        and we can write

d² = 160² / ([tan(25)/tan(20)]² - 1)   Simplifying this, we have

d² = 160² / (.6413959572703699169)     Now....take the square root of both sides

d = 199.7822378461935026188794268 m

And using  ....   h = d*tan(25) ....we have....

h = (199.7822378461935026188794268)* tan(25) ≈ 93.159 m or just  93 m

I think that's it.....I haven't worked one like this before....thanks for submitting it....I hope I haven't made any grevious mistakes !!  (P.S.   ....thanks to alan for catching my previous error.....I went back and corrected it so that now, it appears that I'm smarter than I might really be....also...rom's method is better and actually conforms to what was asked....but I guess my approach shows that there's more than one way to skin a cat !!)   (so to speak)

Let C be the point (0,0,0)

Then D is (-160,0,0)

B is (0,y,0)

A is (0,y,z)

From what we are given

$$\tan(25deg) = \dfrac z y$$

$$\tan(20deg) = \dfrac{z}{\sqrt{160^2+y^2}}$$

solving this we get

z=93.16m which is the the length of AB, i.e. the height of the building

Thanks Chris

Actually...I believe the proper term is "oblique asymptote"....here's a page that might help:

Thank you Chris and Heureka,

I have wondered how to find 'slant' asumtotes (the proper name of these eludes me)

Can anyone show me?

Thursday 19/6/14

Lunch box unwrap (reinout-g)

I believe that the questioner might have also meant this (but I'm not sure).......  y = (x^2+6x-3)/(x-3)

heureka is correct.....there is no horizontal asymptote in a "high/low" rational function.......(I believe we have a "slant" asymptote, instead)

Thanks Kitty.

3x^2 + 4x - 2 = 0     x?

$$\\3x^2+4x-2=0\\\\
3(x^2+\frac{4}{3}x) =2 \quad | \quad :3\\\\
x^2+\frac{4}{3}x =\frac{2}{3}\\\\
\underbrace{x^2+\frac{4}{3}x+ (\frac{4}{2*3})^2}_{
=(x+\frac{4}{2*3})^2
} -(\frac{4}{2*3})^2 =\frac{2}{3}\\\\
(x+\frac{4}{2*3})^2 -(\frac{4}{2*3})^2 = \frac{2}{3}\\\\
(x+\frac{2}{3})^2 -(\frac{2}{3})^2 = \frac{2}{3}\\\\
(x+\frac{2}{3})^2 =(\frac{2}{3})^2 + \frac{2}{3}\\\\$$

$$\\(x+\frac{2}{3})^2 = \frac{2}{3}(\frac{2}{3}+1)\\\\
(x+\frac{2}{3})^2 = \frac{2}{3}*\frac{5}{3} \quad | \quad \pm\sqrt{}\\\\
x+\frac{2}{3} = \pm \sqrt{\frac{2}{3}*\frac{5}{3}}\\\\
x+\frac{2}{3} = \pm \frac{\sqrt{10}}{3}\\\\
x = -\frac{2}{3} \pm \frac{\sqrt{10}}{3}\\\\
x=x_1=\frac{-2+\sqrt{10}}{3}\\\\
x=x_2=\frac{-2-\sqrt{10}}{3}$$

$$x_1=0.38742588672$$

$$x_2=-1.72075922006$$

x^2 + 7x + 10

to factorise this you need to look fr 2 numbers that multiply to +10 and add to +7

Since they mult to a positive it means the signs must be the same and since they add to a pos it means that they are both positive numbers.

2 and 5 are easy to find.

(x+2)(x+5)

This says that we want to subtract f(x) from g(x) and put that result back into f(x).

So g(x) - f(x)  =    (3x + 2) - (2x - 1)    =   x + 3

So  f(x + 3)  =  2(x + 3) - 1  = 2x + 6 - 1  = 2x + 5

x^2 + 7x + 10 = (x+2)(x+5)