it says i need to have at least 20 characters so here you go buddy
Answer:
m ≠1 ( all m in R except 1 )
Step-by-step explanation:
hello :
mx − y + 3 = 0.....(*)
(2m − 1)x − y + 4 = 0 ....(**)
multiply (*) by : -1 you have : -mx+y-3=0 ....(***)
(2m − 1)x − y + 4 = 0 ....(**)
add(***) and(**) : -mx+ (2m − 1)x+1 =0
(2m-m-1)x+1=0
(m-1)x = -1
this system have no solution if : m-1≠0 means : m ≠1
use photo math For that Problem It would work Mate.
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
Answer:
It is A' B'
Pls give me brainliest :)
Step-by-step explanation:
