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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Y=8
your teacher is introducing algebra
Answer:The first table would be a proportional relationship.
Step-by-step explanation:
It's the only table with a constant, which is required in a proportional relationship. The constant would be 1300 which you can find by divided the two numbers in each row.
Answer:
4 4/5
Step-by-step explanation:
Answer: The total cost is
assuming the cost for 1 adult is
and the cost for 1 child is 
Step-by-step explanation:
Assuming the cost for 1 adult is
and the cost for 1 child is
:
and 
Then the expression that gives the total cost
is solved as:
