The answer is y = 3x - 2. To find slope, you subtract y2 - y1 and x2 - x1. In this case, y2 - y1 = 10 - 4 and x2 - x1 = 4 - 2. You would result in 6/2, which can be reduced to 3. Your slope is 3, so the first and second answer automatically are eliminated. Next, you could either draw it out on a coordinate plane or just visualize it. The starting point that goes through each line would be -2, which means D, or y = 3x - 2, is the correct answer.
Answer:
s(t) = -4.9t² + 39.2t
Step-by-step explanation:
Given the projectile motion of an object can be modeled using s(t) = gt2 + v0t + s0, where g is the acceleration due to gravity, t is the time in seconds since launch, s(t) is the height after t seconds, v0 is the initial velocity, and s0 is the initial height. The acceleration due to gravity is –4.9 m/s²
Given
g = -4.9m/s²
v0 = 39.2m/s
s0 = 0m (the initial height)
On substituting into the formula;
s(t) = gt² + v0t + s0
s(t) = -4.9t² + 39.2t + 0
s(t) = -4.9t² + 39.2t
This gives the equation that models the height
I hope this helps you
12/12+x=13/13+26
12/12+x=13/39
12/12+x=1/3
12+x=36
x=24
Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
1/3 = 2/6 = 3/9 = 4/12 = 5/15 = 6/18 = 7/21 = 8/24 = 9/27 = 10/30 = 11/33 = 12/36 = 13/39 = 14/42 = 15/45 = 16/48 = 17/51 = 18/54 = 19/57 = 20/60
You can use any, these are just the smallest to the biggest but any are fine but I would use the first one. Hope I helped. Equivalent is what 3/9 equals to.
