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
Answer: 336g.
Step-by-step explanation:
the formula for density is mass divided by volume (m/v). In this problem, you mus first find the volume of the square block. The formula for volume is length multiplied by width multiplied by height (LxWxH). To solve for volume you plug in the values to the equation: 6x4x2 = 48.
Now, you have the volume, and you have the density. Now you need to solve for mass. Plug in the values you have into the density equation (d=m/v):
7=m/48
to isolate M, multiply each side by 48. this leaves you with 7x48 = m. now all thats left is to solve. 7x48 = 336g. You can double check by plugging the values back into the equation. 336/48 = 7.
Your answer would be
-7-(-i)+6-3i
The greatest common factor is 7m
So, Juan is 3rd, and Tami has at least 2 ppl in front of her, she can't be third because Juan is, so she's fourth. Kari is not 1st, leaving the only space to be 2nd from Kari.