Answer:
Step-by-step explanation:
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.
To learn more about the solution visit:
brainly.com/question/1397278
B is the only number that is divisible by 3
(i) There is glucose outside the Visking tubing. Glucose is small enough to pass through it.
(iii) Normal urine color ranges from pale yellow to deep amber — the result of a pigment called urochrome and how diluted or concentrated the urine is. Pigments and other compounds in certain foods and medications can change your urine color. Beets, berries and fava beans are among the foods most likely to affect the color.
LINK: https://www.mayoclinic.org/diseases-conditions/urine-color/symptoms-causes/syc-20367333
I HOPE THIS HELPS. IF IT DOES, PLEASE MARK BE AS BRAINLIEST. EVERYTHING ELSE IS CORRECT EXEPT (i)
