Solve the inequality 3 ≥ t + 1 question 12 options: a) {t | t ≤ 4} b) {t | t ≤ 2} c) {t | t ≥ 2} d) {t | t ≥ 4}

b

Step-by-step explanation:

ok, lets see the answers one for one.

a) in this case, t can be 4, if t= 4 then t + 1 = 5 and the inequality is false.

b) in this case, the maximum vale of t is 2, then 2 + 1 = 3, the inequality is true here, and all the other numbers in the set is smaller than 2, then all this set keeps the inequality true.

c and d) both sets allow to take, for example, t= 10, wich makes the inequality false.

now in this type of exercises, you should se the biggest and lowest number that keeps the inequality true, in this case the biggest is 2, because 2 + 1 = 3, and there is no lowest limit because any negative number + 1 will be less than 3. With those limits then you can figure the set of valid solutions with all the numbers in between.

the slope of the first function is -3. (3/1)

the slope of the second function is 1. (13-10/6-3).

since |-3| is greater than |1|, the function with the steepest slope is the first.