Dimitristhym

We have 2 numbers $x,y$

$x=4y$ (1)

$x+y=155$<=> We are replacing the (1) => $4y+y=155$ <=> $5y=155$ <=> $y=31$

so we are replacing in the (1) $x=4\times31$<=> $x=124$

so the two numbers? is $31$ and $124$

Hope this helps!

Guest "Exactly one of the axioms of a vector space is not satisfied"

Witch is this axiom? 

Rom yes im sure the exercise say it's not.For this reason i can't solve the exercise

8 if you dont count the square ABCD and 9 if you count it 

Oh ok np then!

I'm fine,you?(Post it like off-topic)

This question its like "we will take the 0" But 0 is not positive so no we will not take it!

Hope I help you to understand​!

Yes right question because the exercise say solutions we will have 4 positive integer values of k

$x= {-b \pm \sqrt{b^2-4ac} \over 2a}={-10\pm \sqrt{10^2-4k^2} \over 2k}$ If  rational solutions means x real so must $10^2-4k^2> 0$

So $k=1,2,3,4$ 

Total 4 positive integer values of k

Hope this helps! 

It's $(3x+4)(9x^2-12x+16)$ because if you do distributive property you will have $27x^3+64$

Hope this helps!