All Answers 358099 Answers

Well!. List all the perfect squares and, by trial and error, find out by subtracting one from the others.

1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Try from the last and go backwards like this:

100 - 81 = 19, which is a prime

100 - 64 = 36, which is NOT prime.

100 - 49 = 51, which is NOT prime......and so on until you reach 1. Then try the next to the last number, or 81 and continue until you finish them all. Then count the pairs that differ by a prime number.And that is it!.

f(f(3)) = 9 - 1 = 8

f(f(8)) = 64 - 1 = 63

I'll do this the bash way

f(2010) = (2011)/(2009)

f(2011/2009) = (4020/2009)/(2/2009) = 4020/2009 * 2009/2 = 2010

You can also notice that f[(x+1)/(x-1)] = x

If I understand your question, here is an example to illustrate it:

P = 21 x 23 x 25 =12,075. The product P =12,075 will always divide P, or: 12,075 / 12,075 = 1 !!!!!.

Since it takes the bus 20 minutes to travel from his home to school and from school to home, that means it takes the bus: 20 / 2 = 10 minutes to travel from home to school or vice versa.

But, when he walks from home to school and takes the bus from school to home, it takes him 40 minutes. That means his walking from home to school, or vice versa, takes him: 40 - 10 = 30 minutes.

Therefore, if he walked both ways, it would take him:
30 x 2 = 60 minutes, or one hour.

how do I do that?

Sorry! I did it two different ways and got 2 different answers!! However, I believe this one is the accurate one.

3.7037037037 E+2018 mod 673 =370

If the points (k, k + 2), (0, k - 6), (k - 4, k) all lie on the same line, find k.

b = 17 - 3*21 = -46.

a=2019;b=(10^a - 1) / 9 * 3;c=c+b; a=a - 1; if(a>=1, goto1, 0);printc

1.8513703703 E+2019 mod 673 = 500

1 / 6^3 =1 / 216

Find the area of the triangle whose vertices are O(0,0), A(6,3) and B(1, -7).

Could you solve with u-substitution?

You must consider the probability that drawing BB, BWB, WWB, or WWBB.  This gives you a probability of 2/9 + 5/81 + 12/81 + 6/81 = 41/81.

We can use a constructive approach, which isolates the unique probabilities for each card.  This gives us an answer of (1/16*c(16,1) + 1/15*c(16,2) + 1/14*c(16,3) + 1/13*c(16,4))/c(16,4) = 27/260.

I know I'm late, but welcome back!!! 

Solve the system \(x^2 = y + 2,\ y^2 = x + 2\).

\(\begin{array}{|lrcll|} \hline (1) & x^2 &=& y+2 \quad \text{or} \quad \boxed{y = x^2-2} \\\\ (2) & y^2 &=& x+2 \\ & \left(x^2-2\right)^2 &=& x+2 \\ & x^4-4x^2+4 &=& x+2 \\ & x^4-4x^2-x+2 &=& 0 \\\\ \boxed{x_1= 2}: & x^4-4x^2-x+2 &\overset{?}=& 0 \\ & 2^4-4*2^2-2+2 &=& 0 \ \checkmark \\ & \text{Polynomial long division}: & \\ & (x^4-4x^2-x+2) &=& (x-2)(x^3+2x^2-1) \\\\ \boxed{x_2= -1}: & x^3+2x^2-1 &\overset{?}=& 0 \\ & (-1)^3+2(-1)^2-1 &=& 0 \ \checkmark \\ & \text{Polynomial long division}: & \\ & (x^3+2x^2-1) &=& (x+1)(x^2+x-1) \\\\ & x^2+x-1 &=& 0 \\\\ & x &=& \dfrac{-1\pm \sqrt{1-4(-1)}} {2} \\ & x &=& \dfrac{-1\pm \sqrt{5}} {2} \\ \boxed{x_3 = \dfrac{-1+\sqrt{5}} {2}} \\ \boxed{x_4 = \dfrac{-1-\sqrt{5}} {2}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && y = x^2-2 \\\\ x_1=1 && y_1 = -1 \\ x_2 =2 && y_2 = 2 \\ x_3 = \dfrac{-1+\sqrt{5}} {2} && y_3 = \dfrac{-1-\sqrt{5}} {2} \\ x_4 = \dfrac{-1-\sqrt{5}} {2} && y_4 = \dfrac{-1+\sqrt{5}} {2} \\ \hline \end{array} \)

Thank you

Squaring both sides as melody said. we get: 

(2x+1)^2=(2x+1)(2x+1)=4x^2+2x+2x+1=4x^2+4x+1 

(4x+3)^2=(4x+3)(4x+3)=16x^2+12x+12x+9=16x^2+24x+9 

So we have

\(4x^2+4x+1=16x^2+24x+9\)

Subtract 1 

\(4x^2+4x=16x^2+24x+8\)

subtract 4x 

\(4x^2=16x^2+20x+8\)

Rearrange 

\(16x^2+20x+8=4x^2\)

Subtract 4x^2

\(12x^2+20x+8=0\)

Factor 4

\(4(3x^2+5x+2)\)=0

\(3x^2+5x+2=0\)

x=\(\frac{-2}{3}\) or -1 

Subsituite back and check 

2(-1)+1=4(-1)+3 

-1 = -1 correct! 

-1 is a solution 

Subsituite -2/3 and check, Correct! 

Here's my solution to 1.

I hope that's right!

If \(a + b = 2\) and \(a^3 + b^3 = 14\), then find all possible values of \(a^2 + b^2\).

\(\begin{array}{|rcll|} \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 + 3ab(a+b) \quad & | \quad a+b = 2,\ a^3+b^3=14 \\ 2^3 &=& 14+ 3ab\times 2 \\ 8 &=& 14+ 6ab \quad & | \quad :2 \\ 4 &=& 7+ 3ab \\ 3ab &=& -3 \\ \mathbf{ab} &=& \mathbf{-1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (a+b)^2 &=& a^2+2ab+b^2 \\ (a+b)^2 &=& a^2+b^2+2ab \quad & | \quad \mathbf{ab=-1} \\ (a+b)^2 &=& a^2+b^2+2(-1) \quad & | \quad a+b = 2 \\ 2^2 &=& a^2+b^2 -2 \\ 4 &=& a^2+b^2 -2 \\ a^2+b^2 &=& 4+2 \\ \mathbf{ a^2+b^2} &=&\mathbf{ 6} \\ \hline \end{array}\)

If one root of (a^2 + 9)x^2 + 13x + 6a = 0 is equal to the reciprocal of the other root, then find a.

\(\begin{array}{|rcll|} \hline (a^2 + 9)x^2 + 13x + 6a &=& 0 \quad & | \quad : (a^2 + 9) \\\\ x^2 + \dfrac{13}{a^2 + 9}x + \underbrace{\dfrac{6a}{a^2 + 9}}_{=r_1r_2} &=& 0 \quad & | \quad \text{vieta} \\\\ r_1r_2 &=& \dfrac{6a}{a^2 + 9} \quad |\quad r_2 = \dfrac{1}{r_1} \\ r_1\dfrac{1}{r_1} &=& \dfrac{6a}{a^2 + 9} \\ 1 &=& \dfrac{6a}{a^2 + 9} \\ a^2 + 9 &=& 6a \\ a^2-6a + 9 &=& 0 \\\\ a &=& \dfrac{6 \pm \sqrt{6^2-4(9)} } {2} \\ a &=& \dfrac{6 \pm \sqrt{0}} {2} \\ a &=& \dfrac{6}{2} \\ \mathbf{a} &=& \mathbf{3} \\ \hline \end{array} \)

a=1;c=3*a^3 + 2*a^2 + a ; if(c%49==3, goto3, goto4);printa,", ",; a++;if(a<=2019, goto1, 0)

Sum n =(19 , 68 , 117 , 166 , 215 , 264 , 313 , 362 , 411 , 460 , 509 , 558 , 607 , 656 , 705 , 754 , 803 , 852 , 901 , 950 , 999 , 1048 , 1097 , 1146 , 1195 , 1244 , 1293 , 1342 , 1391 , 1440 , 1489 , 1538 , 1587 , 1636 , 1685 , 1734 , 1783 , 1832 , 1881 , 1930 , 1979) = 40,959

Circle A and circle B are concentric. A chord of circle A that is tangent to circle B has a length that is twice the radius of circle B. What is the ratio of the area of circle A to the area circle B?

I split the board up into zones.

Zone 1 is the 4 corner squares.  Each of those has 2 adjacent squares

Zone 2 are all the other edge squares. Each of those has 3 adjacent squares

Zone 3 is all the squares in the middle. Each of those has 4 adjacent squares.

I am going to work those out seperately but I will have counted every pair of squares twcie so I have to halve  my answer.

So i get

((36/64)*(4/63)+(24/64)*(3/63)+(4/64)*(2/63))/2 = 0.0277777777777778

This is my best guess.  If you think it is incorrect than please state why you think so.