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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Expression without an equal sign with at least one variable
Hope this helps!
Step-by-step explanation:
State the minimum monthly income and hourly wage per worker needed to cover
monthly expenses for the family you used in part a. Then, explain how to calculate the
hourly wage based on the monthly income and state the hourly wage. Assume that
each full-time worker works four 40-hour work weeks per month, and each part-time
worker works two 40-hour weeks per month. (10 points)
Answer:
Correct option is
A
49m
The circumference of a circular field is 308m
Therefore, 2πr=308
⇒r=
2×22
308×7
⇒r=49m
Step-by-step explanation:
here you go
15.6/ (-3) = -5.2
hope this helps
