Answer:
4th option.
Step-by-step explanation:
PQ is not congruent to QR, instead it is PQ is perpendicular to QR.
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: 1) 2, 3, 4, 9, 32, 279, 8896, 2481705, 22077238784
2) 2,490,924
3) Neither
<u>Step-by-step explanation:</u>
Neither
Not arithmetic because the difference between each of the terms is not the same.
S₂ - S₁ S₃ - S₂ S₄ - S₃
3 - 2 = 1 4 - 3 = 1 9 - 4 = 5
Not geometric because the ratios between each of the terms is not the same.
Answer:
-2.
Step-by-step explanation:
f(0) = 0^2 + 8(0) - 2
= -2.
I am pretty sure the answer is A. 16x+2
