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.
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The answer is x=22 because you just have to add the two equations together and then solve for 8x+4=180
It depends if its decreasing 25% of the original quantity or decreasing by 25% of the quanity that is after already increasing by 25%.
Answer:
idk
Step-by-step explanation:
idk