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:
A
Step-by-step explanation:
Remember that 
x^(b-c)
Using that 
=3^(1/2x-1/2)=3^((x-1)/2)
So we can say: 
, because the bases are the same
We can multiply both sides of the equation by 2. We get x-1=2, and x=3. Which is A.
Answer:
A. Decreasing linear
Step-by-step explanation:
Answer:
I believe the answer is 10 units.
Assuming that you mean 5x(2-3) for your equation, you would just have to plug 2 into where x is. This makes your equation 5*2(2-3). This gives you 10(-1) because 5*2=10, and 2-3=-1. Then simply multiply the two to get:
-10
Hope this helped!
