H=+15 m
v=+5 m/s
Ball hits the ground when h(t)=-15m
h(t)=-9.8t^2+vt+h
=>
-15=-9.8t^2+5t+15
9.8t^2-5t-30=0
Solve for t, using quadratic formula,
t=-1.513 or t=2.023
reject negative root due to context, so
t=2.023 seconds
2)
h(t)=-16t^2+20t+8
a. height before pitch is when t=0, or h(0)=8
b. highest point reached when h'(t)=-32t+20=0 => t=5/8 seconds
c. highest point is t(5/8)=-16(5/8)^2+20(5/8)+8=47/5=9.4 m
d. ball hits ground when h(t)=0 => solve t for h(t)=0
=> t=-0.3187 seconds or t=1.569 seconds.
Reject negative root to give
time to hit ground = 1.569 since ball was pitched.
Slope-intercept form:
y=mx+b
m=slope
b=y-intercept
Data of the first line:
m=-5
b=y-intercept=3 (y-intercept=it is the value of "y" when x=0)
y=-5x+3
A line perpendicular to the line y=mx+b will have the following slope:
m`=-1/m
Therefore: the line perpendicular to the line y=-5x+3 will have the following slope:
m´=-1/(-5)=1/5
Point-slope form of a line: we need a point (x₀,y₀) and the slope (m):
y-y₀=m(x-x₀)
We know, the slope (m=1/5) and we have a point (3,2) therefore:
y-y₀=m(x-x₀)
y-2=1/5(x-3) (point-slope form)
y-2=(1/5)x-3/5
y=(1/5)x-3/5+2
y=(1/5)x+7/5 (slope-intercept form)
Answer: the line perpendicular to the first line will be: y=(1/5)x+7/5
Radius = 15
side = 6.215
a^2 = 15^2 - 6.215^2
a = 13.65
Area = a × p
Area = 13.65 × 99.44
Area = 1357.356
7x+9 equal 10 because it is equal to 5,
Answer:The roots of any quadratic equation is given by: x = [-b +/- sqrt(-b^2 - 4ac)]/2a. Write down the quadratic in the form of ax^2 + bx + c = 0. If the equation is in the form y = ax^2 + bx +c, simply replace the y with 0. This is done because the roots of the equation are the values where the y axis is equal to 0.
Step-by-step explanation:
