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