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:
81 mm²
Step-by-step explanation:
X.x.x (multiplication), it would be x^3
Hope it helped!
we know he made a profit of 1834 for 200 shirts, let's divide those to see how much profit per shirt
so he made a profit of 9.17 per shirt, now profit is surplus value, value beyond the cost, we know its cost was 5.83 per shirt, so if we take 5.83 to be 100%, how much is 9.17 off of it in percentage?
The equivalent sets are <u>X n Y = { } and X U Y= {5, 6, 7,8, 9, 10, 12 ,14}</u>
<h3>Set theory and notation</h3>
Given the following sets expressed as;
X = {8, 10, 12, 14}
Y = {5, 6, 7, 8,9}
Since union means the combination of all the elements in both sets and intersection is the common elements in the set, hence;
X n Y = { }
X U Y= {5, 6, 7,8, 9, 10, 12 ,14}
Note that the intersection is an empty set since there is no common element
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