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28/2/15

27/2/15

Thanks Jilly  

Thanks Bertie, I didn't even think of a bionomial expansion!

Actually Bertie you have got a point. 

I know a lot of old people and very few of them use a cane. 

Many use a frame, that would mean that they have 4 legs and 2 wheels.      

Hence there must be a different answer to the riddle.   

Hi Zegroes,

I missed you   

Thanks Bertie,

I like my mistakes to get picked up.  

If they are not picked up then I am teaching  nonsense and I would never want to do that. 

Thanks Alan,

I've never tried to draw anything so complicated.  

But, I would like to see someone do this algebraically. 

I don't have time and to be honest I don't think I know how.

Do you need to find 3 dimensional gradients ???

This will seriously help you remember it.  

that's interesting Chris

but

Would you like to show us how this property can be shown Chris. 

"But...it can be shown that for any such prime, the exponent 2n would have to be prime,"

Use the change of base rule for each of these

I'll do the first one.

$$log_3 17 = log_{10}17/log_{10}3$$

$${\frac{{log}_{10}\left({\mathtt{17}}\right)}{{log}_{10}\left({\mathtt{3}}\right)}} = {\mathtt{2.578\: \!901\: \!923\: \!162\: \!565\: \!8}}$$

Hi TJM,

I'm sorry but this is not correct.    

$${\frac{{\mathtt{1}}}{\left({\mathtt{2.48}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{3}}}\right)}} = {\mathtt{0.002\: \!48}}$$

This is correct

$$\frac{1}{(2.48*10^3)}=\frac{1}{2.48}\times \frac{1}{10^3} =\frac{1}{2.48}\times 10^{-3}$$

If you are going to do it that way, the thing to do is to square the 1.0275, square that to get 1.0275^4, square that to get 1.0275^8 and finally square again to get 1.0275^16. Then, multiply by the 1.0275^4 obtained earlier. That gets you 1.0275^20 and you can then multiply that by 8435. Can't see why you would want to do that though, unless you're a masochist or looking for some practice in multiplication.

Another possibility is to use the binomial expansion

$$(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+\dots$$

with x = 0.0275 and n = 20. The further you go into the expansion, the smaller the terms become. Keep an eye on the size of the terms to decide how far you want to go.

Call me a wimp, but I think that there are times when it's best to use  a calculator.

Here's a Geogebra visual solution (sorry, I used D for the orthocenter, not H):

.

tan(0+78/60+40/3600)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{0}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{78}}}{{\mathtt{60}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{40}}}{{\mathtt{3\,600}}}}\right)} = {\mathtt{0.022\: \!887\: \!200\: \!781}}$$

$${\mathtt{8\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1.1}}\right)}^{{\mathtt{3}}} = {\mathtt{10\,648}}$$

$$\\FV=Pe^{rt}\\
FV=25,000e^{0.0875*25}\\$$

$${\mathtt{25\,000}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{0.087\: \!5}}{\mathtt{\,\times\,}}{\mathtt{25}}\right)} = {\mathtt{222\,822.574\: \!529\: \!968\: \!375}}$$

$222,822

Hint :Start by multiplying both sides by 20.    That will get rid of all the fractions.

inverse cosine is the same as arc cosine

so use   acos

this is how I would do it.

$$\\8 +\frac{3}{5} - 3 -\frac{4}{5}\\\\
=8- 3 +\frac{3}{5} -\frac{4}{5}\\\\
=5 +\frac{3-4}{5} \\\\
=5 +\frac{-1}{5} \\\\
=4\frac{4}{5}$$

I like these - thanks Chris. 

I already know that this is for a test so i am not going to give you all the answers

$$\\w(x)= \frac{(x+1)}{x}\\\\
$I suggest that you graph it as$\\\\
y=\frac{(x+1)}{x}\\\\
$using Desmos Graphing calculator$\\\\$$

FIRST YOU COUNT THE DECIMAL PLACES. ON THE TOP ROW YOU COUNT 3 PLACES, THEN ON THE BOTTOM ROW YOU COUNT 2 PLACES. SO YOUR FINAL ANSWER YOULL COUNT FIVE PLACES FROM THE RIGHT AND PLACE YOU DECIMAL THERE.

MULTIPLY LIKE THIS:    2005.201

                                x        9050

                               ______________

Perhaps this graph will help:

.

$${\mathtt{881}} \,{mod}\, {\mathtt{21}} = {\mathtt{20}}$$

Theere you go, the remainder is 20