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:

Step-by-step explanation:
At the start of the day there are g pounds of gravel in the quarry.
Then it says that there were 300 pounds added to the mound, so at this point the pounds of gravel in the mound are now g+300.
Then two orders of 700 pounds are sold and removed from the mound. That means they removed 700 pounds from the mound 2 times. Therefore at this point the pounds of gravel in the mound are now g+300-700-700.
Then it says that at this point, the amount of pounds in the mound are 1500. Hence we get

To solve for g we just add 700 two times to both sides of the equation, and subtract 300 from both sides of the equation, getting:

And so 
Answer:
4
Step-by-step explanation:
Plug in 3 as x in the expression:
10 - 2x
10 - 2(3)
10 - 6
= 4
Answer:
Step-by-step explanation:
if additive inverse then
(1/3y)+7
if multiplicative inverse then
3y-7
if both then
3y+7
if in this way
f(x)=(1/3x)-7
f(-x)=(-1/3)-7
Answer:
For x there is no distance since they are both for. For y, they are -6 apart.
Step-by-step explanation: