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:
8>_x>4
>_ this is greater than equal to symbol eight is greater than or equal to x is greater than 4
Answer:
brainly moderators suck
Step-by-step explanation:
Answer:
A=49 degrees
a= 41 degrees
b=90 degrees
Step-by-step explanation:
its asking for the degrees?
Answer:
153.86 feet squared
Step-by-step explanation:
Area of circle = πr²
r is the radius and its already given which is 7 feet so we just plug in the numbers.
area = 3.14*7²
area = 153.86 feet²
