All Answers 358087 Answers

You need to explain how you get the answer anees :/

I'd like someone to point out the flaw in fiora's working. I agree there exist more than one answer, but where is the error in fiora's calculation if the result obtained there is not one of the valid answers?

No it's 12

The best method for dealing with powers and roots of complex numbers, is to switch them to polar form and then to use De Moivre's theorem.

So here,

$$\displaystyle \frac{-7+21\sqrt{3}\imath}{2}=-\frac{7}{2}(1-3\sqrt{3}\imath)=-\frac{7}{2}\sqrt{28}\angle\tan^{-1}(-3\sqrt{3})$$

(ask for details if you need them)

$$\displaystyle = -7\sqrt{7}\angle 280.89339\deg.$$

Raise that to the power one third and you have,

$$(-7\sqrt{7}\angle(-280.89339+360k))^{1/3}$$

$$\displaystyle = -\sqrt{7}\angle(93.63113+120k), \quad k=0,1,2.$$

That gets you

$$\displaystyle k=0:\quad -\sqrt{7}\angle 93.63113 = -\sqrt{7}(\cos93.63113+\imath\sin93.63113)$$

$$\displaystyle =0.16752-2.64044\imath.$$

and the other two values of k,

$$\displaystyle 2.20291+1.46533\imath \;\text{ and } -2.37047+1.17511\imath \quad\text{respectively}.$$

Mmm I'll have to think about this one.   I don't think that i get it Nauseated  

I haven't looked at the question either:((

It is a pity that there are just SO many questions.

It is really hard to find the time to really digest some of the great questions and answers that are given :)

the answear is 2

How many different combinations of quarters, nickels, and dimes can be used to make $0.55 ?

 $$\small{\text{
$
\begin{array}{rccrlclcl}
\hline
\\
1. & \$0.55 &=& & && 3\; Dimes &+& 1\; Quater \\
2. & \$0.55 &=& 1& Nickel && &+& 2\; Quaters \\
3. & \$0.55 &=& 1& Nickel &+& 5\; Dimes \\
4. & \$0.55 &=& 2& Nickels &+& 2\; Dimes &+& 1\; Quater \\
5. & \$0.55 &=& 3& Nickels &+& 4\; Dimes \\
6. & \$0.55 &=& 4& Nickels &+& 1\; Dime &+& 1\; Quater\\
7. & \$0.55 &=& 5& Nickels &+& 3\; Dimes \\
8. & \$0.55 &=& 6& Nickels && &+& 1\; Quater \\
9. & \$0.55 &=& 7& Nickels &+& 2\; Dimes \\
10. & \$0.55 &=& 9& Nickels &+& 1\; Dime \\
11. & \$0.55 &=& 11& Nickels \\
\\
\hline
\end{array}
$}}$$

Ok, I hate to boost his ego still further but TitaniumRome is right.  

You cannot take the square root to simplify a fraction.  

You have to divide top and botton by the same number and since there are no common factors it cannot be simplified.

You can however change it into a mixed numeral :)

what do you want to convert?

Hi TitaniumRome, 

You have to give explanations - they can probably get the answer only from the back of the book.

Your role as an answerer is to help people learn :)

Hi Nauseated,

Yes it was!!  Want another peice?

Bertie is on now, maybe you would like the icecream Bertie ? 

It is of a no melt variety.  Its makers are very ingenious :)

I don't know - maybe you should take your coat off ?

Cost of 1 book = b

$$12b=25.20$$

$$b=\frac{25.20}{12}$$

$${\frac{{\mathtt{25.2}}}{{\mathtt{12}}}} = {\frac{{\mathtt{21}}}{{\mathtt{10}}}} = {\mathtt{2.1}}$$

So one book costs $2.10

Now I have to consider

  $$\\\sqrt[3]{-1}\\
$The first obvious answer is $ -1+0i $ \;\;since (-1)^3=-1\\
$Now, there will be 3 roots$\\
$each will be $2\pi/3 $ radians apart. $\\
$So the other 2 will be $\\
=cos(\pi/3)+isin(\pi/3)\qquad and \qquad cos(-\pi/3)+isin(-\pi/3)\\
=\qquad \frac{1}{2}+\frac{\sqrt3i}{2} \qquad \qquad \qquad and\qquad \qquad \frac{1}{2}-\frac{\sqrt3i}{2}\\
=\qquad \frac{1+\sqrt3i}{2} \qquad \qquad \qquad and\qquad\qquad\frac{1-\sqrt3i}{2}$$

NOW

$$\\\sqrt[3]{-1}=-1,\qquad \frac{1+\sqrt3i}{2}, \qquad\frac{1-\sqrt3i}{2}\\
so\\
\sqrt[3]{14\sqrt[3]{-1}}\\\\
=\sqrt[3]{14(-1)},\qquad \sqrt[3]{14*\frac{1+\sqrt3i}{2}}, \qquad \sqrt[3]{14*\frac{1-\sqrt3i}{2}}\\\\
=\sqrt[3]{-14},\qquad \sqrt[3]{7(1+\sqrt3i)}, \qquad \sqrt[3]{7(1-\sqrt3i)}\\\\
$Now WolframAlpha only gave 3 answers but here I am going to get 3 lots of 3 answers, perhaps each lot is the same as the other 3 lots?$\\\\
\sqrt[3]{-14}\\\\
=\sqrt[3]{14}\sqrt[3]{-1}\\\\
=-\sqrt[3]{14},\qquad \sqrt[3]{14}*\frac{1+\sqrt3i}{2}, \qquad\sqrt[3]{14}*\frac{1-\sqrt3i}{2}\\\\$$

$$\\$This approximates to$\\\\
=-2.410,\qquad 2.410*\frac{1+\sqrt3i}{2}, \qquad2.410*\frac{1-\sqrt3i}{2}\\\\
=-2.410,\qquad 2.410*(0.5+\frac{\sqrt3i}{2}), \qquad2.410*(0.5-\frac{\sqrt3i}{2})\\\\
=-2.410,\qquad 1.205+2.087i, \qquad 1.205-2.087i\\\\$$

Badinage is right.   None of these answers agree with Wolfram|alpha     

how much lek are 129 euro $$\small{\text{$\approx 18000\; \rm{Lek}$}}$$

I had a quick look at Wolfram|Alpha  and  I cannot tell whether it agrees with mine or not since mine is not finished :)

ok then, what about 

$$\\(1+\sqrt3i)^3\\\\
=(\sqrt3i+1)^3\\\\
=[(\sqrt3i)^3]+[3*(\sqrt3i)^2]+[3*(\sqrt3i)]+1\\\\
=(3\sqrt3*-i)+(3*3*-1)+(3*\sqrt3i)+1\\\\
=(-3\sqrt3i)+(-9)+(3\sqrt3i)+1\\\\
=-8$$

SO

$$\\\sqrt[3]{ \frac{14*\sqrt[3]{(1+3i)^3} }{2}}\\\\
=\sqrt[3]{ \frac{14*\sqrt[3]{-8} }{2}}\\\\
=\sqrt[3]{ \frac{14*\sqrt[3]{8}*\sqrt[3]{-1} }{2}}\\\\
=\sqrt[3]{ \frac{14*2*\sqrt[3]{-1} }{2}}\\\\
=\sqrt[3]{ 14\sqrt[3]{-1} } \\\\$$

I have yet to consider   $$\sqrt[3]{-1}$$

Does that look right or wrong to you Badinage?

Melody, if you like doing the Rube Goldberg of math equations, this script will solve it for you.

sum ((39!*(52-i)!)/(40-i)!) /(52!) from i=1 to 40

.

Most trolls are a little deaf, but I hear you loud and clear!

It does seem that I am often behind my time . . . and under my bridge. 

Wolfram alpha doesn't agree with any of us.   

I didn’t have any ice cream, because I was stuffed with your birthday celebration coconut cake. That was delicious!

Sorry Badinage and fiora,  I think you are right - I was suffering from brain freeze

I have not looked from there on Badinage, is the rest of Fiora's work correct?

$$\\(-1+\sqrt3i)^3\\\\
=(\sqrt3i-1)^3\\\\
=[(\sqrt3i)^3]+[3(\sqrt3i)^2(-1)]+[3(\sqrt3i)(-1)^2]+[-1^3]\\\\
=[3\sqrt3*-i]+[-3(3*-1)]+[3(\sqrt3i)]+[-1]\\\\
=[-3\sqrt3i]+[9]+[3\sqrt3i]+[-1]\\\\
=+[9]+[-1]\\\\
=+8$$