<h2>i cant answer it without text sorry. i just cant see the question :(</h2>
The given equation has no solution when K is any real number and k>12
We have given that
3x^2−4x+k=0
△=b^2−4ac=k^2−4(3)(12)=k^2−144.
<h3>What is the condition for a solution?</h3>
If Δ=0, it has 1 real solution,
Δ<0 it has no real solution,
Δ>0 it has 2 real solutions.
We get,
Δ=k^2−144 here Δ is not zero.
It is either >0 or <0
Δ<0 it has no real solution,
Therefore the given equation has no solution when K is any real number.
To learn more about the solution visit:
brainly.com/question/1397278
(2/7)% =
<span>(2 / 7) / 100 = </span>
<span>2 / 700 </span>
<span>2 : 698 </span>
<span>1 : 349</span>
100 times (6+5)
100 times 11
100 groups of 11=1100
Hope I helped:)
