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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Answer:The equation of the line parallel to (3x - y = 7) passes through the point (-5, -3) is y = 3x + 2 and this can be determined by using the one-point slope form.
F(-3) means when x = -3
Find the value on the graph
Solution: f(-3) = 6
Answer:
the answer is d: y=-2x-10
Answer:
4 1/24
Step-by-step explanation:
