All Answers 358093 Answers

Hi Aditya, your answer looks really interesting so I will take a look at your working :)

n belongs to natural no's, then the highest integer m such that 2^m divides 3^2n+2-8n-9 is

=9(9^n)-8n-9

9(8+1)^n  -8n - 9

9(nc08n +nc18n-1 + ...........+ncn80) -8n -9

9(8x+1) -8n-9  (let the remaining terms be expressed as x(ie.nc08n-1,nc18n-2 etc.)

72x+9-8n-9

72x-8n

8(9x-n) divisible by 2m

2m =8

m=3

\(3^{2n+2}-8n-9\\ =9*3^{2n}-8n-9\\ =9*9^n-8n-9\\ =9*(8+1)^n-8n-9\\ =9\left[\begin{pmatrix}n\\0\end{pmatrix}8^0+\begin{pmatrix}n\\1\end{pmatrix}8^1+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}8^n\right]-8^n-9\\ =9\left[1+n*8^1+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+8^n\right]-8^n-9\\ =9\left[8n+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+8^n\right]-8^n\\\)

I think that is correct but I do not think it simplifies to what you have there.......

In response to aditya below.

You want me to factor 8 out, ok I can do that.

\(=9*8\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-8 ^n\\ =2^3*9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^{3n}\\ =2^3*9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^3*2^{3n-3}\\ =2^3*\left[9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^{3n-3}\right]\)

So if n>1 and a natural number then this expression is divisable by 8.  

Note: if n=1 I don't think that this expression IS divisable by 8. because 8^(-1) is a fraction.

Heureka's simple substitution showed this to be incorrect so there must be something wrong  

If you mean in polar form then:

a +b.i = r.e^(i.A)  where r = sqrt(a^2 + b^2)  and A = arctan(b/a)

I just used a piece of numerical software.  I also thought it was a brain teaser at first and looked for a simpler way; however, I couldn't see one! 

Now if the individual terms were squared instead of square-rooted, it would be a different matter.

i mean it s a "brain easer" problem there is need to be easy way

As follows:

how did you calculated this ?

-1+2^1/2-3^1/2+4^1/2-5^1/2............+2014^1/2 ≈ 22.061

You can't square each individual term Solveit.  If x = -1+2^1/2-3^1/2+4^1/2-5^1/2............+2014^1/2

it is not true that x^2 = -1 + 2 -3 + 4 -5 +...-2013 + 2014

-1+2^1/2-3^1/2+4^1/2-5^1/2............+2014^1/2  up to ^2

it will be -1+2-3+4-5......-2013+2014=x^2

x^2=1

x=1

Hallo aditya.

Example: If n = 1 we have:

\(3^{2n+2} - 8n -9 = 3^{2\cdot 1+2} - 8\cdot 1 -9 = 3^4-8-9 = 81-17 = 64 ={\color{red}2^6}\)

\(2^6 \text{ can be devide by } 2^m \text{, if }m=6.\\ So~the~mimimum~ m~ is ~ 6.\\ \text{The maximum m must be equal or greater than the minimum m }.\\ \text{So m is }6\text{ or greater}.\)

m is not a constant.

cool :)

8080*1514 = 12233120 - the score all of the students exept Payton s 

P-     Payton

(12233120+P)/1515=8181

8181*1515-12233120 = 161095

i think the answer is: 161095

\(16x-4-7x\)

\(9x-4\)

9x-4

A helicopter is ascending vertically with a speed of 2.50 m/s. At a height of 145 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

\(\begin{array}{rcl} h &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \qquad | \qquad h = 0\ m \\ 0 &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \\ -\frac12\cdot g\cdot t^2 + v_0\cdot t +h_0 &=& 0\\\\ \boxed{~ \begin{array}{rrcl} ax^2+bx+c &=& 0 \\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\\\ \begin{array}{rcl} a=-\frac12\cdot g\qquad b=v_0 \qquad c = h_0 \\ \end{array}\\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2-4(-\frac12\cdot g)\cdot h_0} \over 2(-\frac12\cdot g)} \\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over -g } \\ t_{1,2} &=& {v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over g } \\\\ \begin{array}{rcl} v_0= 2.50\ \frac{m}{s} \qquad h_0 = 145\ m \qquad g= 9.81\ \frac{m}{s^2}\\ \end{array}\\ t_{1,2} &=& {2.50 \pm \sqrt{2.50^2+2\cdot 9.81\cdot 145} \over 9.81 } \\\\ t_{1,2} &=& \frac{2.50 \pm 149.903802487} { 9.81 } \\ t_1 &=& \frac{2.50 + 53.3961609107} { 9.81 } \\ t_1 &=& 5.69787572994\ s\\\\ t_2 &=& \frac{2.50 - 53.3961609107} { 9.81 } \\ t_2 &=& -5.18819173402\ s \quad \text{ negative time, no solution} \end{array}\)

t = 5,7 s

Answer=x^3+1337x+c

Just hit the "2nd" button.......

{(-1)^0.5}^2={(-1)^2}^0.5

This isn't triue......remember that sqrt(-1)  is defined as " i "   →  i^2   = -1

So....using the order of operations on the left side results in :

[ i ] ^2   = -1

But, using the order of operations on the right results in :

[ 1 ] ^0.5   =  sqrt (1)   = 1

The problem lies in the way the order of operations are performed...... in the first instance, we are squaring a "complex" number, but in the second, we are taking the square root of a real number......

x = 2x + 50   subtract 2x from both sides

-1x = 50      divide through by -1

x = -50

I think you might have this incorrect.....this looks like a kinematic equation for vertical displacement......I believe that it should be:

0.54 = 3.255t  - 4.95t^2     rearrange as :

-4.95t^2 + 3.255t - 0.54   = 0

We can't just divide by t.......

We can use the quadratic formula to solve........however......WolframAlpha shows no real solutions (???)  for t......!!!!

If I've interpeted this incorrectly......sorry....!!!

As an addendum to this problem......let me prove the assertion concerning Euclid....namely that m< TAP = 1/2 minor arc PA

Draw diameter AB

And m< AFB  = m< TAP + m< PAB

And m< AFB = 1/2 arc APB

And m<  PAB = 1/2 minor arc PB

Thus :

m< AFB - m<PAB = m< TAP   and, by substitution

1/2arc APB  - 1/2 minor arc PB  = m<TAP

1/2 [ arc APB - minor  arc PB]  = m< TAP

1/2 [ minor arc PA] = m< TAP

16x - (44 + 77x)    distribute the negative across the terms in the parentheses

16x - 44  - 77x     combine like terms

-61x - 44

1/2 + 3/8    =

4/8 + 3/8   =

7/8