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Imagine you have a large supply of 3kg and 8kg weights.

Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 .

Can you find other combinations of 3kg and 8kg weights whose mean weight is whole number of kg?

What's the smallest?

Whats the largest?

Can you make all the whole numbers inbetween?

\(\begin{array}{|r|r|r|r|} \hline \text{mean} & \text{all combinations} & \text{example combination }~ (n= 1) & \text{example mean }~ (n= 1) \\ \hline 3 & 1\times 3\ kg & \\ 4 & 4\cdot n\times 3\ kg + n\times 8\ kg & 4\cdot 1\times 3\ kg + 1\times 8\ kg & \frac{12+8}{4+1} = 4 \\ 5 & 3\cdot n\times 3\ kg + 2\cdot n\times 8\ kg & 3\cdot 1\times 3\ kg + 2\cdot 1\times 8\ kg & \frac{9+16}{3+2} = 5 \\ 6 & 2\cdot n\times 3\ kg + 3\cdot n\times 8\ kg & 2\cdot 1\times 3\ kg + 3\cdot 1\times 8\ kg & \frac{6+24}{2+3} = 6 \\ 7 & n\times 3\ kg + 4\cdot n\times 8\ kg & 1\times 3\ kg + 4\cdot 1\times 8\ kg & \frac{3+32}{1+4} = 7 \\ 8 & 1\times 8\ kg & \\ \hline \end{array}\)

Thanks, Rarinstraw......that one took me a little while to figure out  how to approach it...!!!!

[ There are  probably easier ways.....but....I couldn't think of any of them....LOL!!!  ]

Thank you CPhill!

2(x+8)^3/4=8     divide both sides by 2

(x + 8)^(3/4)  = 4     raise both sides to the 4th power

(x + 8)^3  = 4^4

(x + 8)^3  = 256  take the cube root of both sides

x + 8  = 4*cuberoot(4)    subtract 8 from both sides

x = 4*cuberoot(4) - 8  = 

4*[ cuberoot(4) - 2]  =

4*[ 2^(2/3) - 2 ]

Average (mean): μ=4.8125
Absolute deviation: 66.875
Mean deviation: 2.08984375
Minimum: 1
Maximum: 10
Variance: 5.96484375
Standard deviation σ=2.44230296032
Corrected sample standard deviation s=2.48138228907
Coefficient of variation cV=0.515611904223
Signal-to-noise ratio SNR=1.93944319712
Median: 4
Mode: {4, 3} - multimodal
Geometric mean: 4.19180456369
Harmonic mean: 3.584
Sum: 154
Sum of absolute values: 154
Average absolute deviation: 2.08984375
Range: 9

Rarinstraw.....here's one way to do this:

Note that the mid-point of the diagonal is located at [ (0 +2)/2, (-3 +3)/2 ] = [2/2 , 0/2] =  (1,0)

Now.....the diagonal length will be =  sqrt [2^2 + (-3-3)^2]  = sqrt(40)

And 1/2 of this  =  sqrt(40)/2

Now....if we construct  a circle centered at (1,0)  with a radius  of 1/2 of the diagonal length, this circle will pass through the given diagonal endpoints as well as the ones we are looking for.  And the equation  of such a circle becomes :

(x - 1)^2  + y^2  = [sqrt(40)/2]^2 

(x - 1)^2 + y^2  = 40/4

(x - 1)^2 + y^2  = 10

Now....the slope of the given diagonal  is  [-3-3] / [0-2]  = -6/-2  = 3

But the other diagonal will lie on a line with the negative reciprocal slope and it will go through the point (1,0)....so the equation  of this line is:

y = (-1/3)(x -1)

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

y = (1 - x)/3

And putting this into our equation of the circle for y to solve for x, we can find the x coordinates of the diagonal endpoints we are looking for :

(x - 1)^2  + ( [x - 1]/3)^2  = 10   simplify

x^2 - 2x + 1  + [1/9](x^2 - 2x + 1)  = 10      multiply through by 9

9x^2 - 18x + 9 + x^2 - 2x + 1  = 90   simplify

10x^2 - 20x - 80  = 0   divide through by 10

x^2 - 2x  - 8  = 0        factor

(x - 4) (x + 2)  = 0      and setting each factor to 0, we have that   x = 4  and x = -2

So.....we can find the y coordinates of the diagonal endpoints, thusly :

When x = 4  ....   y =  [1- 4]/ 3  = -3/3 = -1

When x= -2 ....    y  = [ 1 - (-2)]/ 3  = 3/3  = 1

So....the endpoints of the other diagonal are (4, -1) and ( -2, 1)

Here's a pic that verifies the results :

Solve for x: (2 (1-2 x))/5-1/5 = 15 (2 (1-2 x))/5-1/5 = (2 (1-2 x)-1)/5: (2 (1-2 x)-1)/5 = 15 2 (1-2 x) = 2-4 x: (2-4 x-1)/5 = 15 Add like terms. 2-1 = 1: (1-4 x)/5 = 15 Multiply both sides of (1-4 x)/5 = 15 by 5: (5 (1-4 x))/5 = 5×15 (5 (1-4 x))/5 = 5/5×(1-4 x) = 1-4 x: 1-4 x = 5×15 5×15 = 75: 1-4 x = 75 Subtract 1 from both sides: (1-1)-4 x = 75-1 1-1 = 0: -4 x = 75-1 75-1 = 74: -4 x = 74 Divide both sides of -4 x = 74 by -4: (-4 x)/(-4) = 74/(-4) (-4)/(-4) = 1: x = 74/(-4) The gcd of 74 and -4 is 2, so 74/(-4) = (2×37)/(2 (-2)) = 2/2×37/(-2) = 37/(-2): x = 37/(-2) Multiply numerator and denominator of 37/(-2) by -1: Answer: | | x = (-37)/2

Solve for x: 1/2 (x+8)^3 = 8 Multiply both sides by 2: (x+8)^3 = 16 Taking cube roots gives 2 2^(1/3) times the third roots of unity: x+8 = -2 (-2)^(1/3) or x+8 = 2 2^(1/3) or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: x = -8-2 (-2)^(1/3) or x+8 = 2 2^(1/3) or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: x = -8-2 (-2)^(1/3) or x = 2 2^(1/3)-8 or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: Answer: | | x = -8-2 (-2)^(1/3)      or x = 2 2^(1/3)-8      or x = 2 (-1)^(2/3) 2^(1/3)-8

151

Exponent problems     Maybe???

First divide both sides by 2, then you have    (x+8)^3/4 = 4

Raise both sides to the 4/3 power

((x+8)^3/4)^4/3   =  4^4/3

x+8 = 6.3496

x= - 1.650395

Good night! See you in the morning! :)

Yes, but I need both endpoints, not just one. Thx though Hayley! :D

Summerkim....I think this is supposed to be:

2(x+1/3) = 4 (x-1)     distribute the 2 and 4 across the terms in the parentheses

2x + 2/3  =  4x  - 4        subtract 2x from both sides, add 4 to both sides

4 + 2/3 = 4x - 2x

14/3  = 2x     multiply both sides by (1/2)

(14/3) (1/2)  = (1/2)*2  x

14/6  = x   = 7/3

Well, you have forgotten me! LOL!

Please don't spam. Thank you.

Yeah good point.

One time last year we had about 26 members, but this is still lots of users with:

Why do the teachers say so,lol jk idk