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Well it's simple maths dude..                (a+b)(a-b)=a²-b² when a not equals b..  so...

X+2-2=5-2

x=3

Is there an english version?

Hi KarmaKizuna,

Welcome to Web2.0calc forum, I hope you learn lots here :)

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Make sure you include the link to your question or the post you want them to answer or discuss.

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When you ask a question, it is there for us all to answer, or learn from, so please do not hide it away in a private message :)

Hello Max,

I got the solution to the problem. It is tanθ - tanθ sinθ

I did not know the method used to solve this problem.

Is there any software by which I can can culatate eizen ventor of any martix? I would appreciate yoru help

x=5-2

5-2 = 3

x=3

\(\dfrac{\cos \theta}{\csc\theta +1}\cdot \dfrac{\csc\theta - 1}{\csc\theta - 1}\\ =\dfrac{\cos \theta}{\csc\theta + 1}\\ =\dfrac{\sin 2\theta}{2(1+\sin \theta)}\)

The second fraction just have the same numerator and denominator... 

Hello Heureka,

Pleast check your message center. Thank you for helping me solve the problem. :)

Simplify the following trigonometric expression

Multiply cosθ/cscθ+1 by cscθ-1/cscθ-1

\(\begin{array}{|rcll|} \hline && \left( \frac{ \cos(\phi) } {\csc(\phi)+1 } \right) \cdot \left( \frac{ \csc(\phi)-1 } { \csc(\phi)-1 } \right ) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\csc(\phi)-1 \Big) } { \csc^2(\phi)-1 } \quad & | \quad \csc(\phi) = \frac{1}{\sin(\phi)} \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1}{\sin^2(\phi)}-1 } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1-\sin^2(\phi)}{\sin^2(\phi)} } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { 1-\sin^2(\phi) } \quad & | \quad 1-\sin^2(\phi) = \cos^2(\phi) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos^2(\phi) } \\\\ &=& \frac{ \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(\frac{1-\sin(\phi)}{\sin(\phi)}\Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(1-\sin(\phi)\Big)\cdot \sin(\phi) } { \cos(\phi) } \\\\ &=& \Big(1-\sin(\phi)\Big) \cdot \tan(\phi) \\\\ &=& \tan(\phi)-\tan(\phi)\cdot \sin(\phi) \\ \hline \end{array} \)

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

\(\begin{array}{|rcll|} \hline n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ \hline \end{array} \)

\(\begin{array}{rcll} n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ &=& \frac{1}{\left(\frac{1+2}{2}\right)\cdot 2} +\frac{1}{\left(\frac{1+3}{2}\right)\cdot 3} +\frac{1}{\left(\frac{1+4}{2}\right)\cdot 4} + \ldots + \frac{1}{\left(\frac{1+2013}{2}\right)\cdot 2013} + \frac{1}{\left(\frac{1+2014}{2}\right)\cdot 2014} + \frac{2}{2015} \\ &=& \frac{2}{2\cdot 3} +\frac{2}{3\cdot 4} +\frac{2}{4\cdot 5} + \ldots + \frac{2}{2013\cdot 2014} + \frac{2}{2014\cdot 2015} + \frac{2}{2015} \\ &=& 2\times \Big( \frac{1}{2015} + \frac{1}{2\cdot 3} +\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5} + \ldots + \frac{1}{2013\cdot 2014} + \frac{1}{2014\cdot 2015} \Big)\\\\ && \mathbf{\frac{1}{a\cdot (a+1)} = \frac{1}{a} - \frac{1}{a+1}} \\\\ &=& 2\times \Big[ \frac{1}{2015} + \left(\frac{1}{2}-\frac{1}{3}\right) +\left(\frac{1}{3}-\frac{1}{4}\right) +\left(\frac{1}{4}-\frac{1}{5}\right) + \ldots \\ && + \left(\frac{1}{2013}-\frac{1}{2014}\right) + \left(\frac{1}{2014}-\frac{1}{2015}\right) \Big] \\\\ &=& 2\times \Big[ \frac{1}{2015} + \frac12+ \underbrace{\left(-\frac{1}{3}+\frac{1}{3}\right)}_{=0} +\underbrace{\left(-\frac{1}{4}+\frac{1}{4}\right) }_{=0} +\underbrace{\left(-\frac{1}{5}+\frac{1}{5}\right)}_{=0} + \ldots \\ && + \underbrace{\left(-\frac{1}{2013}+\frac{1}{2013}\right)}_{=0} + \underbrace{\left(-\frac{1}{2014}+\frac{1}{2014}\right)}_{=0} -\frac{1}{2015} \Big] \\\\ &=& 2\times \left(\frac{1}{2015}+ \frac12 -\frac{1}{2015} \right) \\\\ &=& 2\times \left(\frac12 \right) \\\\ &\mathbf{=}& \mathbf{1} \\ \end{array}\)

what is 15% of 157,ooo

Give:  \(157\ 000\ [\times] \ 15\ [\%]\ [=]\)

Result: \(157\ 000\ \times 15 \%=23\ 550\)

or:

\(15\%:x=100\%:157\ 000\)

\(x=\frac{157000\times 15\%}{100\%}\)

\(x=23550\) 

  !

% Error  =    l Estimated - Actual l  /  Actual  =

l 175 -  97.50 l  / 97.50  = 77.50 /  97.50  ≈ .795 = 79.5%

20 / 100  =  .20  = 20%  discount

Yep!!!

One way of doing this is to first convert feet to meters

1 ft  ≈ .3048 m

So.....

12 ft  =  12 * .3048m ≈  3.66m

38 ft =  38 * .3048 m ≈ 11.58

So the total cost =

Area [ in m^2] *  price [per m^2]  =

( 3.66 * 11.58) * 9.99   ≈ 423.4 pounds

I think...your radical needs a  ±  sign in front of it..

\(x=\frac{2\pm2\sqrt3}{2} \\~\\ x=\frac{2(1\pm\sqrt3)}{2} \\~\\ x=1\pm\sqrt3\)               We can reduce this fraction by 2  .

So, your exact solutions are:

x  =  1 + √3        and         x  =  1 - √3

We can get approximate solutions by putting it into a calculator:

x  ≈  2.732         and         x  ≈  -0.732

These should be the  x  values that cause the  y  to equal zero.

Does that help any? I don't know what you mean by  " remove the zero " .

If you want the common log(base 10), then just raise 10 to the power of the log:

N =10^1.059090039

N = 11.4575....

If you want the natural log(ln)(base e), then raise e to the power of your log:

N = e^ 1.059090039

N = 2.8837456988........

I'll perceive this as a compliment :)

Are those dots meant to be degree signs?

If so then

tan^2(45·-30·/2)

\(tan^2(45-30/2)\\=tan^2(45-15)\\=tan^2(30)\\ = (\frac{1}{\sqrt3})^2\\=\frac{1}{3}\)

Pinky people are not here to do all your homework for you.

Be selective about what you want to ask. Learn from your answer!

Tru watching this. 

12x^3+53x^2−34x+5=0

\(\begin{array}{|l|rcll|} \hline & 12x^3+53x^2-34x+5 &=& 0 \\ \mathbf{ x_1 = -5} & 12\cdot (-5)^3 + 53\cdot (-5)^2 -34\cdot (-5) + 5 &=& 0 \\ & -1500+1325+170+5 &=& 0 \\ & 0 &=& 0\ \checkmark \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 12x^2-7x+1 &=& 0 \\ x &=& \frac{7\pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot 1 } }{2\cdot 12} \\ &=& \frac{7\pm \sqrt{1} }{24} \\ &=& \frac{7\pm 1 }{24} \\\\ x_2 &=& \frac{7 + 1 }{24} \\ \mathbf{ x_2 } & \mathbf{=}& \mathbf{ \frac{1 }{3} } \\\\ x_3 &=& \frac{7 - 1 }{24} \\ \mathbf{ x_3 } & \mathbf{=}& \mathbf{\frac{1 }{4}} \\ \hline \end{array}\)

\(x_1 = -5 \\ x_2 = \frac{1 }{3} \\ x_3 = \frac{1 }{4} \)