All Answers 352628 Answers

I will give this a try!

Let total games played =T

Let total games won     =W

Let total games lost       =L

T! /[W! x L!] / 2^T ={[12! ] / [10! x 2!]} / [2^12] =

66 / 4,096 =0.01611328125 x 100 =1.61% - probability of winning 10 out of 12 games played.

Determine all positive integers
\(k \le 2000 \)  
for which
\(x^4+k\)
can be factored into two distinct trinomial factors with integer coefficients.

\(\text{All four roots of $x^4+k$ are complex numbers, because $k$ is a positive integer} \\ \text{The complex numbers are conjugation in pairs :} \)

\(\begin{array}{|rcll|} \hline x^4+k &=& (x-x_1)(x-x_2)(x-x_3)(x-x_4) \quad | \quad x_1,x_2,x_3,x_4 \in C \\ && x_1 = a+bi \\ && x_2 = a-bi \\ && x_3 = u+vi \\ && x_4 = u-vi \\ x^4+k &=& \Big(x-(a+bi)\Big) \Big(x-(a-bi)\Big) \Big(x-(u+vi)\Big) \Big(x-(u-vi)\Big) \\ &=& \Big((x-a)-bi\Big) \Big((x-a)+bi)\Big) \Big((x-u)-vi)\Big) \Big((x-u)+vi)\Big) \\ &=& \Big((x-a)^2-b^2i^2\Big) \Big((x-u)^2-v^2i^2)\Big) \quad | \quad i^2=-1\\ &=& \Big((x-a)^2+b^2\Big) \Big((x-u)^2+v^2)\Big) \\ &=& \Big(x^2-2ax+(a^2+b^2)\Big) \Big(x^2-2ux +(u^2+v^2)\Big) \quad | \quad \text{trinomial factors}\\ &=& x^4-2ux^3+(u^2+v^2)x^2-2ax^3+4aux^2-2ax(u^2+v^2) \\ && +(a^2+b^2)x^2-2u(a^2+b^2)x+(a^2+b^2)(u^2+v^2) \\ &=& x^4 -x^3\underbrace{(2u+2a)}_{=0} +x^2\underbrace{(u^2+v^2+a^2+b^2+4au)}_{=0} \\ && -x\underbrace{\Big(2a(u^2+v^2)+2u(a^2+b^2)\Big)}_{=0} +\underbrace{(a^2+b^2)(u^2+v^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 1. & 2u+2a &=& 0 \quad & | \quad : 2 \\ & u+a &=& 0 \\ & \mathbf{u} &\mathbf{=}& \mathbf{-a} \quad & | \quad \text{or} \quad \mathbf{u^2=a^2} \\ \hline \end{array}\\ &=& x^4 +x^2\underbrace{(a^2+v^2+a^2+b^2-4a^2)}_{=0} \\ && -x\underbrace{\Big(2a(a^2+v^2)-2a(a^2+b^2)\Big)}_{=0} +\underbrace{(a^2+b^2)(a^2+v^2)}_{=k} \\ &=& x^4 +x^2\underbrace{(v^2+b^2-2a^2)}_{=0} -x\underbrace{(2av^2-2ab^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+v^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 2. & 2av^2-2ab^2 &=& 0 \quad & | \quad : 2a \\ & v^2-b^2 &=& 0 \\ & \mathbf{v^2 } &\mathbf{=}& \mathbf{b^2 } \\ \hline \end{array}\\ &=& x^4 +x^2\underbrace{(b^2+b^2-2a^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+b^2)}_{=k} \\ &=& x^4 +x^2\underbrace{(2b^2-2a^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+b^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 3. & 2b^2-2a^2 &=& 0 \quad & | \quad : 2 \\ & b^2-a^2 &=& 0 \\ & \mathbf{b^2 } &\mathbf{=}& \mathbf{a^2 } \\ \hline \end{array}\\ &=& x^4 +\underbrace{(a^2+a^2)(a^2+a^2)}_{=k} \\ &=& x^4 +\underbrace{(2a^2)(2a^2)}_{=k} \\ &=& x^4 +\underbrace{4a^4}_{=k} \\ \hline && \huge{k=4a^4} \\ \hline \end{array}\)

\(\text{trinomial factors:}\)

\(\begin{array}{|rcll|} \hline && \Big(x^2-2ax+(a^2+b^2)\Big) \Big(x^2-2ux +(u^2+v^2)\Big) \\ && \boxed{ b^2 = a^2, \quad u^2 = a^2, \quad v^2=b^2=a^2, \quad u = -a } \\ &=& ( x^2-2ax+ 2a^2)(x^2+2ax + 2a^2) \\ \hline \end{array} \)

Solutions:

\(\begin{array}{|r|r|c|c|} \hline a \in N &k= 4a^4 & k \le 2000 & ( x^2-2ax+ 2a^2)(x^2+2ax + 2a^2) \\ \hline 1 & 4 & \checkmark & ( x^2-2x+ 2)(x^2+2x + 2) \\ 2 & 64 & \checkmark & ( x^2-4x+ 8)(x^2+4x + 8) \\ 3 & 324 & \checkmark & ( x^2-6x+ 18)(x^2+6x + 18) \\ 4 & 1024 & \checkmark & ( x^2-8x+ 32)(x^2+8x + 32) \\ 5 & 2500 & \gt 2000 \text{ no solution } \\ \ldots & \ldots & \gt 2000 \text{ no solution } \\ \hline \end{array}\)

All positive integers k  are 1,2,3,4

The general formula for this that would give the same last 6 digits as 5^131313 is:

5^(16n + 1), where n =1, 2, 3......etc. So, for n = 1, you have: 5^17=762,939,453,125. 

And since 131313 - 1 mod 16 =0, then 5^131313 = will also end in 453, 125.

The only way that  f(x)  + g(x)   is of degree 3  is if the  g(x)  itself is of degree 5   with the first two terms (written in order of descending powers) being -2x^5 + x^4  and the third term being ax^3....where a is a constant.... [there also might be  other terms with  degrees < 3 ]

degree | 5 x^14 + 9 x^13 - 35 x^12 - 15 x^11 - 594 x^10 - 4366 x^8 + 153 x^7 - 21875 x^6 + 9 x^5 - 65691 x^4 + 90 x^3 - 109375 x^2 + 30 x - 78545] =14

A parabola with equation y=ax^2+bx+c has a vertical line of symmetry at x=2 and goes through the two points (1,1) and (4,-1). The quadratic ax^2 + bx +c has two real roots. The greater root is sqrt(n)+2. What is n?

has a vertical line of symmetry at x=2    so     

\(\frac{-b}{2a}=2\\ b=-4a\)

\(y=ax^2+bx+c\\ y=ax^2-4ax+c\)

goes through the two points (1,1) and (4,-1).

\(y=ax^2-4ax+c\\ 1=a-4a+c\\ 1=-3a+c\\ c=1+3a\\~\\ y=ax^2-4ax+1+3a\\ -1=16a-16a+1+3a\\ -2=3a\\ a=-2/3\\~\\ y=\frac{-2x^2}{3}+\frac{8x}{3}-1\\ \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = \frac{\frac{-8}{3} \pm \sqrt{\frac{64}{9}-\frac{8}{3}} }{\frac{-4}{3}}\\ x = \frac{\frac{8}{3} \pm \sqrt{\frac{64}{9}-\frac{8}{3}} }{\frac{4}{3}}\\ x = \frac{\frac{8}{3} \pm \sqrt{\frac{64}{9}-\frac{24}{9}} }{\frac{4}{3}}\\ x = \frac{\frac{8}{3} \pm \sqrt{\frac{40}{9}} }{\frac{4}{3}}\\ x = \frac{8 \pm \sqrt{40} }{4}\\ x=2\pm\frac{\sqrt{10}}{2}\\ \text{The great root is}\\ x=2+\frac{\sqrt{10}}{\sqrt4}\\ x=2+\frac{\sqrt{10}}{\sqrt4}\\ x=\sqrt{2.5}+2\)

n=2.5

factorize   x^-6x-27

I expect you mean  \(x^2-6x-27\)

You have to look for 2 numbers that multiply to -27 and add to -6

Since they mult to a minus number, one must be positive and one must be negative.

Since they add to a minus number the bigger one is the negative one.

-9 and 3 are the numbers

\(x^2-6x-27=(x-9)(x+3)\)

thanks you, i was haveing a very hard time with this problem. Thanks cphill!!!

after two shifts, the vertices go back to the original. the 3rd 100-gon has the same vertices as the first. the sum is 2009. sorry I am bad at explaining, but I know for sure the answer is 2009

Here's my best effort....

Since a checker is placed on a diagonal....we have 8  possibilities  for the checker's placement

If we place the checker on Row 1, Column 1   we have  64 total checkers  = 8^2

If we pllace the checker on Row2, Column 2, we have 49 total checkers  = 7^2

Following this pattern, the expected value is

[8^2 + 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2 ] / 8  =   204 / 8 =  51/2 =  25.5 checkers

there are 8 choices for the first checker...idk if it matters

Note that  in each  "piecewise"  part of the function, the  greatest value of  [f(x)  - x]  occurs at the left part of the interval  and is   0

And note that the least value of  f(x)  - x  occurs at the right part of the piecewise interval and is  "almost"  -1

So...the range of  f(x)  - x   is  ( -1, 0 ] 

I'm not sure about this, but here's my take

Note that any two elements drawn from the  set   { 60, 61, 62, 63,......,98, 99, 100}  will always have a  sum > 120  

And  the set  {1, 2, 3,.....,17, 18, 19}   can be combined with the first set such that no two elements drawn from this "combined" set will have a sum of 120

So...the sum of the elements of this set will be :

(19)(20) / 2    +   [100 + 61] * 40/2 + 60

190  + 161* 20 + 60   =

190 + 3220 + 60 =

3470

3.Find the equations for two lines that both go through the point (1,2), one with slope 3 and one with slope -1/3. (write the numerical coefficient of each term to complete the required equation) 
the answer would be.
a. With the slope =3: __ x-y- ____=0  (my answer y=3x-1)
b. With slope = -1/3: x + ___ y - ___=0 (my answer y=-1/3x+7/3)

First line

y = (3)(x - 1) + 2

y = 3x - 3 + 2

y = 3x  -  1      subtract y from both sides and rearrange as

3x - y - 1  = 0   [ your answewr is correct in slope-intercept form ]

Second line

y= (-1/3) ( x - 1) + 2     multiply through by 3

3y = (-1)(x - 1) + 6

3y  = -x + 1 + 6

3y  = -x + 7       add x to both sides, subtract 7 from both sides

x + 3y  - 7   = 0

2.Find the equation of the lines that are both passing through the point (10,-8) with slopes 5 and -1/5. 
Answer: ___ x - y - ____ = 0  (  my answer y=5x-58)
x + ___ y+ ____ = 0   (my answer y=-y=-1/5x-6)

First line

y = 5(x - 10) +  -8

y = 5x - 50 - 8

y = 5x - 58       subtract y from both sides

5x - y - 58  = 0     [  this is the same as your answer, just in "standard form" ]

Second line

y  = (-1/5) ( x - 10) + -8       multiply through by 5

5y  = (-1)(x - 10)  +  - 40  simplify

5y  = -x + 10  - 40

5y  = -x - 30      add x, 30 to both sides

x + 5y  + 30  = 0  [ your answer is in slope-intercept form and is correct.....I think they want "standard form" ]

1.Find the equation of the line which contains the point (-2,3) and is parallel to the line 3x+5y=17. (write the numerical coefficient of each term to complete the required equation) 
Answer would be: ___x + ___ y - ___ = 0 

Let us find the slope of the given line...in the form Ax + By  = C, the slope will be given by   -A/B

So...the slope of the given line  =  -3/5

And since parallel lines have the same slope, we can  write an equation of a line having a slope of -3/5 and passing through ( -2,3) thusly

y  =(-3/5)(x -  -2)  + 3      simplify

y = (-3/5) ( x + 2)  + 3      multiply through by 5

5y  = -3(x + 2) + 15

5y = -3x - 6 + 15

5y = -3x + 9     add 3x to  both sides, subtract 9 from both sides

3x + 5y - 9 = 0

Well...it  COULD be a right triangle with legs of 6950  and 3760  [ the hypotenuse will not be an integer value]

If so....the area  =    (1/2) (product of the legs)   = (1/2)(6950)(3760)  = 13726250 cm^2

Let us assume that  we can factor  x^4 + k  in the following manner :

(x^2  + ax  + b) (x^2 - ax  + b)

Expanding this, we have 

x^4 + ax^3 + bx^2

      -  ax^3 -a^2x^2 - abx

                  +bx^2   + abx   + b^2

___________________________

x^4      +   (2b - a^2)x^2         + b^2

Since the  x^2 term will "disappear", this implies that  b^2  = k

Then  b can possibly range from 1 - 44   since 1^2  = 1  and 44^2  = 1936.....note that 45^2  = 2025 which is too large

And since the  x^2 term disappears, it must be that  2b - a^2  =  0

Add a^2 to both sides  and we have that   2b  = a^2

Taking the square root of this we have that

√[2b]  = a

Since a is an integer, then  √[2b]  must also  be an integer

And the values that make √[2b]  an integer is when  b  = 2    b  = 8    b = 18   and b  = 32

So  (a, b)   =  ( 2,2)  (4, 8)   ( 6, 18)    and  ( 8, 32)

So........the following  product trinomials are possible

( x^2 + 2x + 2) ( x^2 - 2x + 2)  =  x^2 + 4

(x^2 + 4x + 8) (x^2 - 4x + 8) = x^2  + 64

(x^2 + 6x + 18) (x^2 - 6x + 18)   x^2 + 324

(x^2 + 8x  + 32) (x^2 - 8x + 32)  = x^2 + 1024

I think you need more information about your triangle, since it doesn't appear to be right-angled triangle. You must know either 2 sides and one angle, or two angles and one side....etc. in order to calculate the area.

5^131313 =   70 498 418 681 817 ..................................... 942 169 189 453 125

=7.0498418................x 10^91,783

Did you even read the question?

I think that the easiest way to prove that \((3n)^2+(4n)^2=(5n)^2\) has infinitely many solutions is to solve for n:
 

This is incorrect, those terms do not cancel out.