So, the right option is 1. so if my ans was helpful u can follow me.
The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
<u>Explanation:</u>
We need to find the time at which the ball will be at height 19 feet.
Equation:
h = 3 + 34t - 16t²
19 = 3 + 34t - 16t²
16 = -16t² + 34t
-16t² + 34t - 16 = 0
On solving the equation, we get
t1 = 0.7 s and t2 = 1.42s
Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
Slope-intercept form: y = mx + b
(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)
For lines to be parallel, they have to have the same slope.
y = 6x + 6 The slope of this line is 6, so the parallel line's slope is also 6.
Now that you know m = 6, substitute/plug it into the equation:
y = mx + b Plug in 6 for "m" in the equation
y = 6x + b To find "b", plug in the point (20, 1) into the equation
1 = 6(20) + b
1 = 120 + b Subtract 120 on both sides to get "b" by itself
1 - 120 = 120 - 120 + b
-119 = b Now that you know b = -119, plug it into the equation
y = 6x - 119
To find the slope, we put this in y = mx + b form where m is ur slope.
3x - 4y = 7.....subtract 3x from both sides
-4y = -3x + 7 ...now we divide both sides by -4
(-4/-4)y = (-3/-4)x + (-7/4)...simplify
y = 3/4x - 7/4
y = mx + b
y = 3/4x - 7/4.....so the number in the m position is 3/4 <== ur slope
