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The following grid shows a magic square.  
What is the sum of the three numbers in any row?
\(\begin{array}{c|c|c} 2x & 3 & 2 \\ \hline & & -3 \\ \hline 0 & x & \end{array}\)

Find x:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} 2x & 3 & 2 \\ \hline & & -3 \\ \hline 0 & x & \color{red}y \end{array} \\ 0+x+y &=& 2-3+y \\ x &=& 2-3 \\ \mathbf{x}&=& \mathbf{-1} \\ \hline \end{array}\)

The sum of the three numbers in any row:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} \color{red}-2 & \color{red}3 & \color{red}2 \\ \hline & & -3 \\ \hline 0 & -1 & \end{array} \\ -2+3+2 &=& \mathbf{3} \\ \hline \end{array}\)

Find y:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline & \color{red}y& -3 \\ \hline 0 & -1 & \end{array} \\ 0+y+2 &=& 3 \\ y &=& 3-2 \\ \mathbf{y}&=& \mathbf{1} \\ \hline \end{array}\)

Find y:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline & 1& -3 \\ \hline 0 & -1 & \color{red}y \end{array} \\ -2+1+y &=& 3 \\ -1+y &=& 3\\ y &=& 3+1\\ \mathbf{y}&=& \mathbf{4} \\ \hline \end{array}\)

Find y:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline\color{red}y & 1& -3 \\ \hline 0 & -1 & 4 \end{array} \\ -2+y+0 &=& 3 \\ y &=& 3+2 \\ \mathbf{y}&=& \mathbf{5} \\ \hline \end{array}\)

The magic square:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline 5 & 1& -3 \\ \hline 0 & -1 & 4 \end{array} \\ \hline \end{array} \begin{array}{rcll} 0+1+2 &=& 3 \\ -2+3+2 &=& 3 \\ 5+1-3 &=& 3 \\ 0-1+4 &=& 3 \\ -2+1+4 &=& 3 \\ -2+5+0 &=& 3 \\ 3+1-1 &=& 3 \\ 2-3+4 &=& 3 \\ \end{array}\)

\(\frac{6a^3}{b}:\frac{ab^2}{12}\qquad \qquad b\ne0\\ \frac{6a^3\div a}{b}:\frac{ab^2\div a}{12}\\ \frac{6a^2}{b}\times 12b:\frac{b^2}{12}\times 12b\\ 72a^2:b^3\\ \)

You can continue to rearrange this. It depends on what form you want it in.

Edit: I have corrected a small error. 

We construct the following sequence:
Beginning at 0, we move to the next whole number by changing our first number by one,

our second number by two, our third number by three, etc.
If we are able to move backwards to a whole number that has not been previously listed, we do so, otherwise we move forward.
The first five numbers of this sequence are 0, 1, 3, 6, 2.
Find the 20th number in this sequence.
What is the first number to appear twice in this sequence?

Recaman's sequence:

\(\begin{array}{|r|r|} \hline 1 & 0, \\ & & +1 \\ 2 & 1, \\ & & +2 \\ 3& 3, \\ & & +3 \\ 4& 6, \\ & & -4 \\ 5& 2, \\ & & +5 \\ 6& 7, \\ & & +6 \\ 7& 13, \\ & & +7 \\ 8& 20, \\ & & -8 \\ 9& 12, \\ & & +9 \\ 10& 21, \\ & & -10 \\ 11& 11, \\ & & +11 \\ 12& 22, \\ & & -12 \\ 13& 10, \\ & & +13 \\ 14& 23, \\ & & -14 \\ 15& 9, \\ & & +15 \\ 16& 24, \\ & & -16 \\ 17& 8, \\ & & +17 \\ 18& 25, \\ & & +18 \\ 19& 43, \\ & & +19 \\ \color{red}20& \color{red}62, \\ & & -20 \\ 21& \color{green} 42, \\ & & +21 \\ 22& 63, \\ & & -22 \\ 23& 41, \\ & & -23 \\ 24& 18, \\ & & +24 \\ 25& \color{green} 42, \\ & & -25 \\ 26& 17, \\ & &+26 \\ 27& 43, \\ \ldots & \ldots \\ \hline \end{array}\)

Thanks very much Alan,

It took me a while to work out what you were saying but I have it all sorted now.

I never would have thought to do that!  

Have you worked it out mathmathj28?

If so what answer did you finally get?

In a University out of 120 students, 15 opted mathematics only, 16 opted statistics only, 9 opted physics only and
45 opted physics and mathematics, 30 opted physics and statistics, 8 opted mathematics and statistics, and
80 opted physics.
Find the sum of number of students who opted mathematics and those who didn't opted any of the subjects given.

My attempt:

\(\begin{array}{|rcll|} \hline x+y+t+9 &=& 80 \quad | \quad x+y = 45 \\ 45+t+9 &=& 80 \\ t+54 &=& 80 \\ t &=& 80-54 \\ \mathbf{ t } &=& \mathbf{26} \\ \hline \end{array} \begin{array}{|rcll|} \hline 30 &=& y+t \\ -~~8 &=& y+z \\ \hline 22 &=& y+t-(y+z) \\ 22 &=& y+t-y-z \\ 22 &=& t-z \\ z+22 &=& t \\ \mathbf{ z } &=& \mathbf{t-22} \quad | \quad \mathbf{ t =26} \\ z &=& 26-22 \\ \mathbf{ z } &=& \mathbf{4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline z+y &=& 8 \\ y &=& 8-z \quad | \quad \mathbf{z=4} \\ y &=& 8-4\\ \mathbf{ y } &=& \mathbf{4} \\ \hline \end{array} \begin{array}{|rcll|} \hline x+y &=& 45 \\ x &=& 45-y \quad | \quad \mathbf{y=4} \\ x &=& 45-4\\ \mathbf{ x } &=& \mathbf{41} \\ \hline \end{array}\)

The sum of number of students who opted mathematics:

\(\begin{array}{|rcll|} \hline \text{Mathematics} &=& 15+x+y+z \\ \text{Mathematics} &=& 15+41+4+4 \\ \mathbf{\text{Mathematics}} &=& \mathbf{64} \\ \hline \end{array} \)

The sum of number of students who didn't opted any of the subjects given:

\(\begin{array}{|rcll|} \hline \text{didn't opted any} &=& 120-(9+15+16+x+y+z+t) \\ \text{didn't opted any} &=& 120-(40+41+4+4+26) \\ \text{didn't opted any} &=& 120-115 \\ \mathbf{\text{didn't opted any}} &=& \mathbf{5} \\ \hline \end{array}\)

Let f(n) be the sum of the positive integer divisors of n. If n is prime and f(f(n)) is also prime, then call n a bouncy prime. What is the smallest bouncy prime?

3 is prime.  It has positive integer divisors 1 and 3, so f(3) = 4.  

4 has divisors 1, 2 and 4, so f(4) = 7 which is prime.

Hence n = 3 is a bouncy prime.

This doesn't hold for n = 2, so n = 3 is the smallest bouncy prime.

Consider g(1) = f(1)^16  This will consist of the sum of the even degree coefficients and the sum of odd degree coefficients..  Now consider g(-1) = f(-1)^16.  This will consist of the sum of the even degree coefficients and the negative of the sum of the odd degree coefficients.  Add g(1) and g(-1).  This will eliminate the odd degree coefficients, but will leave twice the sum of the even degree coefficients. Hence to get the sum of the even degree coefficients you need:

                                                (g(1) + g(-1))/2  or  (f(1)^16  +  f(-1)^16)/2

I would like to see someone present this answer too.  

Your function f certainly works but I do not know what to do next either.

Find the area of the quadrilateral with vertices (0,3), (3,0), (0,9), and (9,0).

A(0,3)      B(3,0)      C(9,0)      D(0,9)

BC = 6          angle DCB = 45º

Area of the quadrilateral (ABCD) = (9² - 3²) / 2  

21/100  of  35 = 21/100 * 35

you finish it!

Take a photo of what you have tried, and show it to us.

Another hint.

The radius of a circle always stays the same.  That is what the pair of compases is for.

Two points A and B are chosen at random on a circle of radius 1.  What is the probability that AB is greater than 1?

I think that no explanation is needed.  

you have to bisect it

but i cant get iit right

Hint

You will need to use a pair of compasses.

Two points A and B are chosen at random on a circle of radius 1.  What is the probability that AB is greater than 1?

The lengths of the sides of a triangle are 5, 6, and 10 cm respectively. Find the square of the length of the median of the greatest side.

a = 5           b = 6        c = 10

(m_c)² = {[ sqrt( 2a² + 2b² - c² )] / 2}²  = 5.5 cm²  

See Melody's reply to your earlier post.

During a football game, Matt kicked the ball three times.
His longest kick was 43 metres and the three kicks averaged 37 metres.
If the other two kicks were the same length, the distance, in metres, that each travelled was what?

\(\begin{array}{|rcll|} \hline \dfrac{43+2x}{3} &=& 37 \\\\ 43+2x &=& 37*3 \\ 43+2x &=& 111 \\ 2x &=& 111-43 \\ 2x &=& 68 \\\\ 2x &=& \dfrac{68}{2} \\\\ \mathbf{ x } &=& \mathbf{34}\ \text{metres} \\ \hline \end{array}\)

See https://web2.0calc.com/questions/math-a-bit-of-help-needed#r1

The terms of a particular sequence are determined according to the following rules: * If the value of a given term is an odd positive integer $s$, then the value of the following term is $3s - 9$ * If the value of a given term is an even positive integer $t$, then the value of the following term is $2t - 7$. Suppose that the terms of the sequence alternate between two positive integers $(a,b,a,b,\ldots)$. What is the sum of the two positive integers?

The lengths of the sides of a triangle are 5, 6 and 10 cm respectively.

Find the square of the length of the median of the greatest side.

cos-rule:

\(\begin{array}{|lrcll|} \hline (1) & 6^2 &=& 5^2+10^2-2*5*10*\cos(A) \\ & 36 &=& 125-100\cos(A) \\ & 100\cos(A) &=& 125-36 \\ & 100\cos(A) &=& 89 \\ &\mathbf{ \cos(A) } &=& \mathbf{\dfrac{89}{100}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (2) & x^2 &=& 5^2+5^2-2*5*5*\cos(A) \\ & x^2 &=& 50-50\cos(A) \\ & x^2 &=& 50-50*\dfrac{89}{100} \\ & x^2 &=& 50-\dfrac{89}{2} \\ & x^2 &=& 50-44.5 \\ &\mathbf{ x^2 } &=& \mathbf{5.5} \\ \hline \end{array}\)

The square of the length of the median of the greatest side is \(\mathbf{5.5}\ \text{cm}\)

The lengths of the sides of a traingle are c=5, b=6 and a=10 cm respectively. Find the square of the length of the median of the greatest side.

Hello Guest!

\(m_a^2=\frac{1}{4}\cdot (2(b^2+c^2)-a^2)\\ m_a^2=\frac{1}{4}\cdot (2(36+25)-100)cm^2\)

\(m_a^2=5.5cm^2\)

  !

People should not be doing your assignment for you. And you should not be requesting it.

Could you please explain how you got there? Thanks

A square is inscribed in the ellipse \(x^2 + 4y^2 = 4\), with two side parallel to the x-axis.  
What is the area of the square.

\(\text{Let the side of the square $s$}\)

\(\begin{array}{|rcll|} \hline x^2 + 4y^2 &=& 4 \quad | \quad x=y=\dfrac{s}{2} \\ \left(\dfrac{s}{2}\right)^2 + 4*\left(\dfrac{s}{2}\right)^2 &=& 4 \\ 5*\left(\dfrac{s}{2}\right)^2 &=& 4 \\ \left(\dfrac{s}{2}\right)^2 &=& \dfrac{4}{5} \\ \dfrac{s^2}{4} &=& \dfrac{4}{5} \\ \mathbf{s^2} &=& \mathbf{\dfrac{16}{5}} \\ \mathbf{s^2} &=& \mathbf{3.2} \\ \hline \end{array}\)

The area of the square is \(\mathbf{3.2}\)