All Answers 358149 Answers

Well there is of course but working out the best cuts would take more time than i have right now.

If you let the side lengths of the equilateral triangle be 2 units then the height will be $$\sqrt3$$  units.

You can multiply these by any constant if you want different side lengths.

The area of the triangle will be      $$A=(1/2)*b*h = 0.5*2*\sqrt3 = \sqrt3$$

The are of a square is    $$l*l$$  so

$$\\l*l=\sqrt3\\
l^2=3^{1/2}\\
l=3^{1/4}\qquad$or if you prefer $ l=\sqrt[4]{3}\;\; units$$

so if the side of the triangle is 2k units then the side of the square will be   $$\sqrt[4]{3}\times k\;\; units$$

Mmm,

first  $$\frac{\alpha}{2}$$   is between  45 and 67.5 degrees.   so that is the first quadrant.   All trig ratios are positive

so   $$\alpha$$    must be between 90 and 135 degrees so that is the second quadrant.  Only sine and cosec are positive all other ratios will be negative,

Now

$$\\cos \left(\frac{\alpha}{2}\right)=\frac{5}{8}\\\\
cos^2 \left(\frac{\alpha}{2}\right)=\frac{25}{64}\\\\
cos^2 \left(\frac{\alpha}{2}\right)+sin^2 \left(\frac{\alpha}{2}\right)=1}\\\\
\frac{25}{64}+sin^2 \left(\frac{\alpha}{2}\right)=1}\\\\
sin^2 \left(\frac{\alpha}{2}\right)=1-\frac{25}{64}\\\\
sin^2 \left(\frac{\alpha}{2}\right)=\frac{39}{64}\\\\
sin \left(\frac{\alpha}{2}\right)=\frac{\sqrt{39}}{8}\\\\\\$$

$$\\sin(\alpha)
= sin(\frac{\alpha}{2}+\frac{\alpha}{2})\\\\
sin(\alpha)=2sin\frac{\alpha}{2}cos\frac{\alpha}{2}\\\\
sin(\alpha)=2*\frac{\sqrt{39}}{8}*\frac{5}{8}\\\\
sin(\alpha)=\frac{\sqrt{39}}{4}*\frac{5}{8}\\\\
sin(\alpha)=\frac{5\sqrt{39}}{32}\\\\\\
Cosec(\alpha)=\frac{32}{5\sqrt{39}}\\\\
Cosec(\alpha)=\frac{32\sqrt{39}}{5*39}\\\\
Cosec(\alpha)=\frac{32\sqrt{39}}{195}\\\\$$

$$\\cos(\alpha) = cos(\frac{\alpha}{2}+\frac{\alpha}{2})\\\\
cos(\alpha)=cos^2(\frac{\alpha}{2})-sin^2(\frac{\alpha}{2})\\\\
cos(\alpha)=\frac{25}{64}-\frac{39}{64}\\\\
cos(\alpha)=-\frac{14}{64}\\\\
cos(\alpha)=-\frac{7}{32}\\\\\\
sec(\alpha)=-\frac{32}{7}\\\\\\$$

$$\\tan(\alpha)=sin(\alpha)}\div{cos(\alpha)}\\\\
tan(\alpha)=\frac{5\sqrt{39}}{32}\div\frac{-7}{32}\\\\
tan(\alpha)=\frac{5\sqrt{39}}{32}\times\frac{32}{-7}\\\\
tan(\alpha)=-\frac{5\sqrt{39}}{7}\\\\\\
cot(\alpha)=-\frac{7}{5\sqrt{39}}\\\\
cot(\alpha)=-\frac{7\sqrt{39}}{5*39}\\\\
cot(\alpha)=-\frac{7\sqrt{39}}{195}\\\\$$

The fraction that are children =

1 - (1/4) - (3/5)  =

20//20 - 5/20 - 12/20  =

3/20

Great anon except that A can be positive or negative :)

(4/3)pi(13502)^3/(4/3)pi(4455.5)^3

like this - you just need to add in the * signs

(4/3)*pi*(13502)^3/(4/3)*pi*(4455.5)^3

$${\frac{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{13\,502}}\right)}^{{\mathtt{3}}}}{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{4\,455.5}}\right)}^{{\mathtt{3}}} = {\mathtt{2\,148\,737\,683\,594\,873\,814\,584\,912.192\: \!948\: \!379\: \!139\: \!779}}$$

If you do this in the home page calc you can press equals a second time and it will present it in scientific notation.  

$$\sqrt{-\pi} = i\sqrt{\pi}$$

What does the # mean ?

$$\left({\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{17}}}\right){\mathtt{\,\times\,}}\left({\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{28}}}\right)$$

= $${\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left({\mathtt{17}}{\mathtt{\,-\,}}{\mathtt{28}}\right)}$$

= $${\mathtt{56}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{11}}\right)}$$

= $${\mathtt{5.6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{11}}}$$

= $${\mathtt{5.6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{11}}\right)}$$

=  $${\mathtt{5.6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{10}}}$$

you mean estimate it on the web2 calc?  You just need to add the * signs :)

60*sqrt(31)-80*sqrt(17)

6-1×0+2÷2

=6-0+1

=7

[2nd] [acos]

arc cos is the same as inverse cos

$${\frac{{\mathtt{62.5}}}{{\mathtt{100}}}} = {\frac{{\mathtt{5}}}{{\mathtt{8}}}} = {\mathtt{0.625}}$$

I could do that in my head Chris but then I am a show off.  LOL

Here is another easier way and it works for all numbers ending in 5

25^2=2^2+2 then add 25 = 625

75^2 = 7^2+7 then add 25 = 5625

35^2=3^3+3 then add 25 = 1225

105^2=10^2+10 then add 25 = 11025

etc.

That is really easy to do in your head :)

Another great answer Brittany 

nice work BJ.

I have put your answer aside for the next wrap :)

Please can another mathematician check this answer. 

I'm not good at limits so my answer will probably be wrong but I always think I learn better if I get involved.

Lim x-> infinity of (e^x + x)^1/x

$$\\\displaystyle\lim_{x\rightarrow\infty}\;(e^x+x)^{1/x}\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\left(\frac{e^x}{1}\cdot \frac{(e^x+x)}{e^x}\right)^{1/x}\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\left[\left(1+ \frac{x}{e^x}\right)^{1/x}\cdot (e^x)^{1/x}\right]\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\left[\left(1+ \frac{x}{e^x}\right)^{1/x}\cdot e\right]\\\\
=\;e\times\;\left[\left(1+ 0}\right)^{1/x}\right]\\\\
=\;e$$

The amount is given by

15000(1.06)^33   .....I let you plug this into the calculator to get the answer.....

well the web2 calc will solve equations for you.

e.g

i enter

x^2-7=8             And the calc gives me the answer.  Isn't that nifty :)

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{7}} = {\mathtt{8}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\sqrt{{\mathtt{15}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{15}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{3.872\: \!983\: \!346\: \!207\: \!416\: \!9}}\\
{\mathtt{x}} = {\mathtt{3.872\: \!983\: \!346\: \!207\: \!416\: \!9}}\\
\end{array} \right\}$$

The radius - r -  of an octagon with side length s can be figured thusly :

(1/2)s / sin 22.5  = r   =  s/(2sin22.5)

And the total area is gven by

A = 8*(1/2)r^2sin45 .... and substituting for r, we have

A = 4s^2/[2sin22.5]^2 * sin 45

A =  s^2 [sin45/[sin22.5]^2]  

So...if the side length is 7 times as great....the area =

A = (7s)^2  [sin45/[sin22.5]^2] = 49s^2 [sin45/[sin22.5]^2]  = 49 times as large = 49 x 35 = 1715 sq units

https://www.youtube.com/watch?v=O8ezDEk3qCg

19 computers and 23 peripheral graphics devices

I don't get this question.

The connections are not 1 to 1 since there are more peripheral devices than computers.

So some computers are connected  to more than 1 device. 

What is the maximum number of divices that a computer can connect to?

Can a peripheral connect to more than one computer?

Se my confusion.   

The answer could be really enormous depending on what the question really means .

Perhaps you mean the  [2nd] key ?

I disagree Chris,

For me the  answer would be 0.  Why would I put my hand into a bag full of sharp little tacks - that would be crazy!