Idk
At least 20 characters to explain it well.
y = 0.6x - 3.2
The general form of the desired equation is
y = mx + b
where
m = slope of the line
b = y intercept of the line
If two lines are parallel, their slopes will be the same, Since the slope of the
given line "y = 0.6x +3" is 0.6, that will also be the slope of the desired line.
So our equation becomes:
y = 0.6x + b
Now we can substitute the x and y value of the desired point we want the new line to pass through and find b. So
y = 0.6x + b
-5 = 0.6(-3) + b
-5 = -1.8 + b
-3.2 = b
So the desired equation is now
y = 0.6x - 3.2
Rx - sx + y = b
WHEN SOLVING FOR X :
rx - sx + y = b
We must get x onto it's own side, so subtract y from both side.s
rx - sx = b - y
Then, factor out x.
x(r - s) = b - y
Then, divide both sides by (r - s).
x(r - s) ÷ (r - s) = b - y ÷ (r - s)
Simplify.
x = b - y / r - s →

WHEN SOLVING FOR Y :
rx - sx + y = b
We need to isolate y, so get rid of everything BUT y on the left side.
Subtract rx from both sides.
-sx + y = b - rx
Then, add sx to both sides.
y = b - rx + sx
~Hope I helped!~
Answer:
a. h = 60t − 4.9t²
b. 12.2 seconds
c. 183.7 meters
Step-by-step explanation:
a. Given:
y₀ = 0 m
v₀ = 60 m/s
a = -9.8 m/s²
y = y₀ + v₀ t + ½ at²
h = 0 m + (60 m/s) t + ½ (-9.8 m/s²) t²
h = 60t − 4.9t²
b. When the ball lands, h = 0.
0 = 60t − 4.9t²
0 = t (60 − 4.9t)
t = 0 or 12.2
The ball lands after 12.2 seconds.
c. The maximum height is at the vertex of the parabola.
t = -b / (2a)
t = -60 / (2 × -4.9)
t = 6.1 seconds
Alternatively, the maximum height is reached at half the time it takes to land.
t = 12.2 / 2
t = 6.1 seconds
After 6.1 seconds, the height reached is:
h = 60 (6.1) − 4.9 (6.1)²
h = 183.7 meters