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Thanks, yeah I think i was just overcomplicating it trying to think of some rule that would always work relating the points involved, but in the end it's just equating the 2 equations lol.

so  🔵=x

🔴=y &

⚪=z

in equation form it is:

(x*y*z)+(x*y*z)+(x*y*z)=y*y*y which we can simplyfy to:

3(x*y*z)=y*y*y which means that

x*y*z=(y³)/3 so

x=(y²) / (3*z)

so we can choose any random numbers for y & z and put them into the formula to work out x. if you want to to work out y it's

y=sqrt(x*3*z)

so we can choose any number for x & z then work out y from that. or for z it's

z=((y²)/x)/3

 i have checked these and they always work. so there is an infinite number of possible solutions. and just out of interest; CPhill says that one answer could be that x=1 y=4 and z=8 but (1*4*8)+(1*4*8)+(1*4*8)=32+32+32=96 but y*y*y=4*4*4=64 so CPhill's answer is saying that 64=96 which is not true. 

\((x+ay)(x+ay)=(x+ay)^2\)

And remember that:

\((a+b)^2 = a^2+2ab+b^2\)

So using this rule, we can just substitute a = x and ay = b

\(x^2+2xay+ay^2\)

You can use the rule to skip the process of using the distributive rule :)

Which set of ordered pairs represents a function?

a. (5, 4), (4, 5), (1, 2), (1, 4)

b. (2, 5), (1, 6), (0, 5), (1, 10)

c. (0, 0), (1, 1), (2, 0), (3, 3)

d. (2, 1), (3, 1), (5, 1), (5, 4)

The most basic thing to remember is that ever x vlaue has to have AT MOST ONE y value to go with it.

in a)     if x is 1, y can be 2 or 4 so that is definitely not a function!

You should be able to decide aobut the others now :)

Purplemath, for maths

https://www.purplemath.com/

Hi Kryn,

Sorry about the answering delay :/

You will get the hang of these types of equation rearranging quite easily I think. 

It just takes a bit of practice.

The idea is always to keep the 'equation balanced' 

So you decide what will help you get to the end that you want and then you have to ALWAYS do the same thing to both sides.

So say you want to get from 

\(y+1=-\frac{4}{3}(x-5)\)

to

\(y-3=-\frac{4}{3}(x-2)\)

Lets take a look  

\(y+1=-\frac{4}{3}(x-5)\)

First I want y-3 on the left so if I subtract 4 on the left that will work.  BUT I must do the same on the right side.

\(y+1-4=-\frac{4}{3}(x-5)-4\\ y-3=-\frac{4}{3}(x-5)-4\\ y-3=-\frac{4}{3}(x-5)-\frac{4*3}{3}\\ y-3=-\frac{4}{3}(x-5)-\frac{4}{3}*3\\ \mbox{Factorize out the }\frac{-4}{3}\\ y-3=-\frac{4}{3}(x-5+3)\\ y-3=-\frac{4}{3}(x-2)\\ \)

I could also have expanded the bracket as the 3rd step and then collected like terms and factorize the -4/3 again afterwards.  The answer would  have been the same.

I hope that helps but by all means aske more questions if you want more help or clarification.  

----------------------------------------------

When you sove equations you are 'undoing' calculations so  you do it in the opposite order to BODMAS.  :)

C.

For A, y = -5/x, the x is in the denominator of the fraction. so it's not a  linear equation.

For B, y = 2 - 5x², that's obviously a quadratic equation.

For D, y = -5x², that's obviously a quadratic equation.

B.

For A. f(x)= 2x², That's obviously a quadratic.

For C, f(x)= 1 / 2x, the x is in the denominator of the fraction, not the numerator, so it's not a linear function.

For D, f(x)= - 1/ 2(x-2), the x is in the denominator of the fraction, not the numerator, so it's not a linear function neither.

So the answer is B

i think it's c. y = 2 - 5x

If you insist on a formal numerical method here's one way of doing it:

but this is a very heavy-handed way to solve such a simple problem!

Hi Melody and Chris, 

thank you.

I have calculated myself on the top line.
Sorry!

Hello Max!

\(\text{What's } (x+\frac{1}{x})^4 \text{ ?}\) 

\( (x+\frac{1}{x})^4 \)  = \((x^{2} + \frac{x}{2} + \frac{1}{x^{2} })^{2} \) 

^{\(= x^{4} + \frac{x^{3} }{2}+ 1 + \frac{x^{3} }{2}+\frac{x^{2} }{4} + \frac{1}{2x}+ 1 + \frac{1}{2x}+ \frac{1}{x^{4} } \)}

^{\(= x^{4}+ x^{3} + 2+ \frac{x^{2} }{4} + x+ \frac{1}{x^{4} } \)}

Greeting asinus :- )

   !

Hi Heureka, 

Chris and I were discussing just this morning that we wished you would come back.

Were your ears burning?

It is really GREAT to see you again  :)

It's simple -3 - 5 = -8

Remember that -*+ = -

x=i, -i 

What's

\((x+\frac{1}{x})^4 ~ \text{ ?}\)

\(\begin{array}{rcll} && (x+\frac{1}{x})^4\\ &=& \binom40 \cdot x^4 + \binom41 \cdot x^3 \cdot \frac1x + \binom42 \cdot x^2 \cdot \frac1{x^2} + \binom43 \cdot x \cdot \frac1{x^3} + \binom44 \cdot \frac1{x^4} \\ &=& 1 \cdot x^4 + 4 \cdot x^3 \cdot \frac1x + 6 \cdot x^2 \cdot \frac1{x^2} + 4 \cdot x \cdot \frac1{x^3} + 1 \cdot \frac1{x^4} \\ &=& x^4 + 4 \cdot x^2 + 6 + \frac4{x^2} + \frac1{x^4} \\ \end{array} \)

Her loss is $75 and her benefit is $175 so her total benefit is:

Profit-Loss = Pure Profit

$175-$75 = $100 

Can you solve it using Raphson method?

Why there are 2 of the same post help LOL

\(f(x)=4x^2+3x-2\\ f(-3)= 4(-3)^2+3(-3)-2=25\\ .\\ .\\ f(x+1)=4(x+1)^2+3(x+1)-2\\= 4x^2+8x+4+3x+3-2\\=4x^2+11x+5\)

\(f(2)+f(-2)\\ = 4(2)^2+2(2)-2+4(-2)^2+2(-2)-2\\ = 16 + 4 - 2 + 16 -4 -2 \\ = 28\)

Good job guest, for that table.

P.S. I am using an alt LOL