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Thank you so much Alan ^^ I'll try to read the books that you recommended

thanks for your answer ^^

thanks Emily! I just want to know how it can be useful ^^

Oh ok Heureka, I can see what you mean :)

Hello Melody,

i am not certain,

if      \(z =-i\)

then  \(z^2 = (-i)\cdot(-i) = (-i)^2 = i^2 = -1\)

if      \( z^2 = -1 \)

then \(z = \sqrt{-1}= i\), but z can also be -i

                                 this  is not equivalent (<=>)

Hello Melody,

i am not certain,

if z =-i then z² = -1

if z² = -1 then z = \(\sqrt{-1}\) = i, but z can also be -i

7.6*105%=7.98

7.98*35=279.3

Guest,

It is very rude of you to re-ask your question when it has already been answered.  

You should always say thank you to the answerer and if you do not like the answer then you say why you do not like the answer.

Give the original answerer feedback and also five him/her the oportunity to fix it if there is an error or explain more if you need it.

Always remember your manners !

Thanks for your answers Chris and EightMersennePrime   :))

what is the root of root of 2? And why? 

\(\sqrt{(\sqrt2)}=(2^{0.5})^{0.5}=2^{0.25}=\sqrt[4]{2}\)

Hi Heureka,

I certainly accept that your knowledge of mathematics is far superior to mine but I do not get why I cannot do this.

Our answers are the same. 

I suppose it is some mathematics nuance that has gone over my head :(

But thank you   

two curves have equations y=x^2-4x+7 and y=6x-x^2-1.

Find the area between the two curves

\(\begin{array}{|rcll|} \hline f\left(x\right)=6x-x^2-1 \\ g\left(x\right)=x^2-4x+7 \\ \hline \end{array}\)

intersections:

\(\begin{array}{|rcll|} \hline g\left(x\right) &=& f\left(x\right) \\ x^2-4x+7 &=& 6x-x^2-1 \\ \dots \\ 2x^2-10x+8 &=& 0 \quad & | \quad : 2 \\ x^2-5x+4 &=& 0 \\ (x-1)(x-4) &=& 0 \\ x_1 = 1 &\text{and}& x_2 = 4\\ \hline \end{array}\)

area between the two curves:

\(\begin{array}{|rcll|} \hline && \int \limits_{1}^{4}(~ f(x) - g(x) ~)\ dx \\ &=& \int \limits_{1}^{4}(~ 6x-x^2-1 - ( x^2-4x+7 ) ~)\ dx \\ &=& \int \limits_{1}^{4}(~ 6x-x^2-1 - x^2+4x-7 ~)\ dx \\ && \dots \\ &=& \int \limits_{1}^{4}(~ 10x-2x^2-8 ~)\ dx \\ &=& [~ 10\frac{x^2}{2}-2\frac{x^3}{3}-8x ~]_1^4 \\ &=& [~ 5x^2 -\frac{2}{3}x^3-8x ~]_1^4 \\ &=& [~ 5\cdot (4)^2 -\frac{2}{3}(4)^3-8\cdot 4 ~]-[~ 5(1)^2 -\frac{2}{3}(1)^3-8\cdot 1 ~] \\ &=& 5\cdot 16 -\frac{2}{3}\cdot 64-32 -5 +\frac{2}{3}+8 \\ &=& 51 -\frac{2}{3}\cdot 64 +\frac{2}{3} \\ &=& 51-\frac{2}{3}(64 - 1) \\ &=& 51-\frac{2}{3}(63) \\ &=& 51-2\cdot 21\\ &=& 9 \\ \hline \end{array} \)

The area between the two curves is 9

me neither

Hello Melody,

you set -1 to \(i^2\) but i think this is not allowed, if (-1) is a complex number.

only (-1) real is i^2 and (-1) complex is \(e^{i\pi}\).

This way is okay \(i^2 \rightarrow -1\)
This way is not allowed \(-1 \rightarrow i^2\)

(-81)^\frac{1}{4}

\(\begin{array}{|rcll|} \hline x&=& \sqrt[4]{-81} \\ &=& (-81)^\frac{1}{4} \\ &=& (81)^\frac{1}{4} \cdot (-1)^\frac{1}{4} \\ &=& 3 \cdot (-1)^\frac{1}{4} \quad & | \quad e^{i\pi} = -1\ ! \quad (\text{Euler's Formula for Complex Numbers}) \\ x_1 &=& 3\cdot (e^{i\pi})^\frac{1}{4} \\ &=& 3\cdot e^{i\frac{\pi}{4}} \\ &=& 3\cdot [\cos(\frac{\pi}{4}) + i\cdot \sin(\frac{\pi}{4})] \quad & | \quad \cos(\frac{\pi}{4})=\sin(\frac{\pi}{4})= \frac{\sqrt{2}}{2} \\ &=& 3\cdot (\frac{\sqrt{2}}{2}+ i\cdot \frac{\sqrt{2}}{2} ) \\ &=& \frac{3 \sqrt{2}}{2}(1+i) \\\\ x_2 &=& 3\cdot e^{i(\frac{\pi}{4}+\frac{\pi}{2} )} \\ x_2 &=& 3\cdot e^{\frac{3}{4} i \pi } \\\\ x_3 &=& 3\cdot e^{i(\frac{\pi}{4}+2\cdot\frac{\pi}{2} )} \\ x_3 &=& 3\cdot e^{\frac{5}{4} i \pi } \\\\ x_4 &=& 3\cdot e^{i(\frac{\pi}{4}+3\cdot\frac{\pi}{2} )} \\ x_4 &=& 3\cdot e^{\frac{7}{4} i \pi } \\\\ \hline \end{array} \)

Use your brain then, it works for me.

I forgot to finish it............this is one answer but there are 3 more...

I see Heureka is answering so I expect he will give the other ones.  :)

Thanks in advance Heureka  

How do you put a quadratic into vertex form when you have an a value?

Lets see

\(y=ax^2+bx+c\\ y-c=ax^2+bx\\ y-c=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)-a(\frac{b}{2a})^2\\ y-c+a(\frac{b}{2a})^2=a(x^2+\frac{bx}{a}+(\frac{b}{2a})^2)\\ y-c+(\frac{b^2}{4a})=a(x+\frac{b}{2a})^2\\ \frac{1}{a}\left(y+(\frac{b^2}{4a}-c)\right)=(x+\frac{b}{2a})^2\\ (x+\frac{b}{2a})^2=\frac{1}{a}\left(y+(\frac{b^2}{4a}-c)\right)\\ \text{So the vertex is} \left( \frac{-b}{2a},(c-\frac{b^2}{2a}) \right) \)

Pretty easy.....just multiply every element by 3  =

12     -6

  3     12

You did fine, EMP.....you answered this :

[x^2-x-30] / [ x-6] = 8      [which is what I would have assumed ].....but....

Since he/she said no brackets....I assumed the literal form  of

x^2 - x -30/x - 6 = 8

Either way....they have a choice of two presentations.....LOL!!!!

Taken literally...we have

x^2-x-30/x-6 = 8

x^2 - x -30/x -14  = 0   multiply everything by x

x^3 - x^2 - 30 - 14x  = 0

x^3 - x^2 - 14x - 30  = 0

By the Rational Roots Theorem, 5 is a root....and we can use synthetic division to find the remaining polynomial / roots

5 [  1     - 1      -14       -30  ]

               5       20        30

      1       4         6          0

The remaining polynomial  is :

x^2 + 4x  + 6    ......this has no real roots......using the quad formula, the other two roots are :

-2 ± i√2

No problem, I was just wondering.

You should create an account 

Hello CPhill,

i think this is a good method to check the calculation.

Perfect thank you so much, sorry for the double post, just working out how the site works. Great site

I thought I already answered it, did I answer it wrong or something?