The way I figured this one out was saying that if it takes 2lbs to stretch it to 12 inches and 5lbs to stretch to 18 inches that means that 3 additional pounds = 6 inches. Therefore, each additional pound is 2 inches. We know already that with 2 lbs it was 12 inches in length and we can also say that 2 lbs = 4 inches. So subtract 4 from 12 and you end up with the answer of 8 inches.
We can get the answer to this equation by solving for the variable (r).
7.4 - 9 - 2r = 18
We want to get r on one side, so we want to push everything else to the other side.
-2r = 18 - 7.4 + 9
-2r = 19.6
2r = -19.6
r = -9.8
What do you want exactly i don’t understand the question. Comment and ill solve it
B. Is the right answer
First you have to take the common elements then use an identity/formula to get the rest
x^3 - 3x^2 + x-3
x^2 (x-3) +1 (x-3)
(x^2 +1) (x-3)
(x-1)(x+1)(x-3) {using a^2-b^2 on x^2-1^2}
Answer:
t ∈ {1, 3}
Step-by-step explanation:
You want to find t such that ...
h = 27
27 = -8t^2 +32t +3 . . . . . . substitute the expression for h
24 = -8t^2 +32t . . . . . . . . . subtract 3
-3 = t^2 -4t . . . . . . . . . . . . . divide by -8
1 = t^2 -4t +4 = (t -2)^2 . . . . add 4 to complete the square
±√1 = t -2 . . . . . . . . . . . . . . take the square root
t = 2 ± 1 . . . . . . . . . . . . . . . . add 2
t = 1 or 3
The object is 27 ft off the ground at t = 1 and again at t = 3.