Answer : C
hope that helps
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
Please enclose the "nt" inside parentheses: <span>A(t)=P(1+r/n)^(nt).
Then: A = $500*(1+0.06/12)^(5*12) = $641.68</span>
Answer:
The means differ by 1, but the ranges differ by 40.
Step-by-step explanation:
The mean for LaTesha's score is (92+45+67+36+80)/5= 64
The mean for Benards score is (63+68+62+69+53)/5= 63
The range for LaTeshas score is 92-36=56
The range for Benards score is 69-53=16
So, 64-63=1 and 56-16= 40
Answer:
7 notepads
Step-by-step explanation:
5 + 3x = 26
Subtract constant (5)
3x = 21
Divide by coefficient (3)
x=7
