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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brainly.com/question/1397278
The denominator can't be equal to 0.
The domain is all numbers except -3 and 3.
Answer:
22 in
Step-by-step explanation:
<u>circumference of a circle = 2 π r</u>
where radius (r) = 7/2 in. = 3.5 in
C = 2 π (3.5)
= 22 in
Answer:
is the last one
Step-by-step explanation:
Answer:
See below ↓
Step-by-step explanation:
a. 4 ÷ (2/3) = 12/3 ÷ 2/3 = 12 ÷ 2 = 6
b. 4 ÷ 1 = 4
Attachment below.
