For which values of t with t ∈ r, the equation has no solutions:

The equation will have no solution for 6 > t > - 2.

Step-by-step explanation:

The quadratic equation ax² + bx + c = 0 will have imaginary or no solution when the discriminant i.e. b² - 4ac < 0

Now, in our case the equation is x² - tx + t + 3 = 0

So, condition for imaginary or no solution is (-t)² - 4(1)(t + 3) < 0

⇒ t² - 4t - 12 < 0

⇒ t² - 6t + 2t - 12 < 0

⇒ (t - 6)(t + 2) < 0

So, either (t - 6) > 0 and (t + 2) < 0

⇒ t > 6 and t < - 2 which is not possible.

or, (t - 6) < 0 and (t + 2) > 0

⇒ t < 6 and t > - 2

Therefore, the equation will have no solution for 6 > t > - 2. (Answer)

range is infinite on both x and y axis positive and negative

step-by-step explanation:

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead