All Answers 359411 Answers

But I'm willing to be set straight!

NB: The last expression still reduces to e^-5, so L'Hopital's rule still works of course, but I see it as an unnecessarily complicated approach here.

.

(arctanx)' = ?

$$\tan{ (\arctan{(x)} ) } = x \quad | \quad \frac{d}{dx} \\\\
\left[ 1 + \tan^2{( \arctan{(x)} )} \right] \times [\arctan{(x)} ]' = 1
\right(
\\\\
\left( 1 + x^2 \right) \times [\arctan{(x)} ]' = 1
\right)\\\\
\boxed{[\arctan{(x)} ]' = \frac{1}{ 1 + x^2 }
\right)}$$

But we didn't study this method 

We always try to simplify it ti give us zero by zero or infinity by infinity then using l hopital rule 

In this case L'Hopital's rule makes things ten times worse!

.

Mmmmm....OK....

Sorry, Melody......I mucked it up originally by reading ".25" as "25"..............you understood it as I did....

(arctanx)' = ? (arcsinx)' = ?

$$(\arcsin(x))' = \ ? \\\\
\boxed{ \sin{ ( \arcsin{(x)} ) } = x } \quad | \quad \frac {d}{dx} \\\\
cos{ ( \arcsin{(x)} ) } * (\arcsin(x))' = 1$$

$$\small{
(\arcsin(x))' =
\frac{1}
{ cos{ ( \arcsin{(x)} ) }
} \quad | \quad cos{ ( \arcsin{(x)} ) } = \sqrt{ 1- sin^2{ ( \arcsin{(x)} ) } } = \sqrt{ 1- x^2 } }
}\\\\
\boxed{(\arcsin(x))' = \frac{1}
{ \sqrt{ 1- x^2 } }}$$

When y=2

$$4x^2=1+4/9$$

$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{9}}}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{6}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{6}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.600\: \!925\: \!212\: \!577\: \!331\: \!5}}\\
{\mathtt{x}} = {\mathtt{0.600\: \!925\: \!212\: \!577\: \!331\: \!5}}\\
\end{array} \right\}$$

width at the top =  $${\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{3}}}}$$    metres

It appears to me that we all got different answers.  

I for one did not really understand the question.  

Now Chris's is the same as mine (he did a little correction) but Heueka's is still different.

Maybe Heurka  interpreted it differently from me and Chris?

We have

(x^2 / .25) - (y^2 / 9 )  = 1

And, let's assume that the base of the pillar lies on the line y = -2

Then, since the pillars are 4 m high, at the top of the pillar, y = 2

So we have

x^2 / .25  - 2^9 / 9 = 1

x^2 / .25 - 4/9  = 1        add 4/9 to each side

x^2 / .25  = 1 + 4/9

x^2 / .25  = 13 / 9        multiply by .25 on both sides

x^2 = (.25 * 13) / 9

x^2 = 13/36

x = ±√(13) / 6

And these are the two x values on the hyperbola when y = 2

So.....the distance between these two x values is the width of the cross-section at the top of the pillar =  (2)√(13)/6 = (1/3)√(13) =  aboiut 1.2019 m

An architect's design for a building includes some large pillars with cross sections in the shape of hyperbolas. The curves are modeled by . If the pillars are 4 meters tall, find the width of the top of each pillar.

$$\dfrac{x^2}{0.25} - \dfrac{y^2}{9} = 1 \quad \text{we need x}\\\\
\dfrac{x^2}{0.25} = 1 + \dfrac{y^2}{9} \\\\
x^2 = 0.25*(1 + \dfrac{y^2}{9}) \quad | \quad \pm\sqrt{} \\\\
x_{1,2} = \pm 0.5 * \sqrt{ (1 + \dfrac{y^2}{9} ) } \\ \\
x_{1,2} = \pm \dfrac{ 0.5}{3} * \sqrt{ y^2 + 9} \\ \\
x_{1,2} = \pm \dfrac{ 1}{6} * \sqrt{ y^2 + 9} \\ \\
x_1= -\dfrac{ 1}{6} * \sqrt{ y^2 + 9} \\\\
x_2= \dfrac{ 1}{6} * \sqrt{ y^2 + 9}$$

$$width = x_2 - x_1 = \dfrac{ 1}{6} * \sqrt{ y^2 + 9 }- \left(-\dfrac{1}{6} * \sqrt{ y^2 + 9 } \right) \\\\
width = 2*\dfrac{1}{6} * \sqrt{ y^2 + 9 } \\\\
width = \dfrac{1}{3} * \sqrt{ y^2 + 9 } \quad | \quad y = 4\ meter\\\\
width = \dfrac{1}{3} * \sqrt{ 16 + 9 } \\\\
width = \dfrac{1}{3} * \sqrt{ 25 } \\\\
width = \dfrac{1}{3} * 5 \\\\
width = \dfrac{5}{3} \\\\
width = 1.\overline{6} \ m$$

√125

Your question is a little ambiguous. Is this what you mean?:

xπ = -1.3127494544162674

x = -1.3127494544162674/π

x = -0.4178611294230754

∴ -1.3127494544162674 = -0.4178611294230754π

$${{\mathtt{5}}}^{-{\mathtt{3}}} = {\frac{{\mathtt{1}}}{{\mathtt{125}}}} = {\mathtt{0.008}}$$ 

That's because (imagine n is any number): n^-e= 1/n^e

So for example: 6^-2 = 1/6^2 or 1/36

Well I can demonstrate with a pattern

4*3    =12

4*2     =8

4*1     =4

4*0     =0

4*-1    =-4

4*-2    =-8

Can you see that the pattern shows that a pos * a neg = a neg  ?

Now i will swap it around

-2*4   =-8

-2*3  =-6

-2*2  =-4

-2*1  =-2

-2*0  =0

-2*-1  =+2

-2*-2  =+4

-2*-3   =+6

Can you see how the pattern shows that a neg * a neg = a pos  ?

there are other ways to show this as  well.  One of the other answerers may show you another way. :)

$${\frac{\left({{\mathtt{10}}}^{{\mathtt{13}}}\right)}{\left({{\mathtt{10}}}^{{\mathtt{10}}}\right)}} = {\mathtt{1\,000}}$$

1000 years

You could use your photo if you want to. :)

I assume that is you?  You are a pretty girl   

Because that's the way it is. See? That is what happens.$${\frac{\left(-{\mathtt{45}}\right)}{\left(-{\mathtt{54}}\right)}} = {\frac{{\mathtt{5}}}{{\mathtt{6}}}} = {\mathtt{0.833\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

x=-y/3+1

x-1=-y/3

3x-3=-y

y=-3x+3

this is the inverse function :)

Are you sure it is correct?

It looks like some financial formula but it does not look right.

Is it a part of a bigger problem?

It looks like it will be horrible to solve as it is now.  

Yes it is but from here it is the problem.

An excellent answer - thank you. :)

$$\frac{1.90(1.20)}{(1 + R)} + \frac{1.90(1.20)(1.15) }{ (1 + R)^2} + \frac{1.90(1.20)(1.15)(1.10)}{(1 + R)^3} +\frac{ \frac{1.90(1.20)(1.15)(1.10)(1.05) }{ (R-0.05) }}{(1 + R)^3 }= $34.02$$

I am not answering right now but is this your intended question?

(I believe it is what you have asked)

Thanks Alan - I didn't know that 'by definition' bit.  

Actually you haven't used L'hopitals rule though ?  

did you really just did that..........a stupid mistake for a fifth grader!