Find all values of k for which the equation 3x2−(k+2)x+k−1=0 has no solutions.

there are solutions for all values of k

Step-by-step explanation:

In order for there to be no real solutions, the value of the discriminant must be negative. For this equation, the discriminant is ...

(-(k+2))^2 -4(3)(k-1)

We want to find values of k for which this is negative:

(k^2 +4k +4) -12k +12 < 0

k^2 -8k +16 < 0

(k -4)^2 < 0

A square is never negative, so there are no values of k that result in the equation having no solutions.

answer: .75 km

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!