C(12/15) because the other 3 all equal 0.666666666666667 but 12/15 equals 0.8
We are given a graph for the height of the ball with respect to horizontal distnace x of the ball from Julie.
From the graph, we can see the position of Julie is at (0,0). And starting height of the ball when thrown is 6 feet.
Then it moved to the maximum height of 14 feet. And hit to the ground when it has position on x-axis is 3.5 feet.
<em>Because we can see the parabola is crossing x-axis at 3.5 on the right.</em>
<em>That represents distance of the ball from Julie when it hit to the ground.</em>
<h3>Therefore, it has a horizontal distance of 3.5 feet from Julie.</h3>
Hey there!!!
What is slope-intercept form ?
<em>The slope intercept form is y = mx + b </em>
Where 'm' is the slope and 'b' is the y-intercept.
Given equation :
... 4x - 4y = 15
This is given is standard form.
Let's convert this into slope-intercept form.
... 4x - 4y = 15
... -4y = -4x + 15
... y = x + ( 15 / 4 )
<em>Hence, the slope is '1'. </em>
Hope my answer helps!
I agree with the other person ^^^
Answer:
Here's a quick sketch of how to calculate the distance from a point P=(x1,y1,z1)
P
=
(
x
1
,
y
1
,
z
1
)
to a plane determined by normal vector N=(A,B,C)
N
=
(
A
,
B
,
C
)
and point Q=(x0,y0,z0)
Q
=
(
x
0
,
y
0
,
z
0
)
. The equation for the plane determined by N
N
and Q
Q
is A(x−x0)+B(y−y0)+C(z−z0)=0
A
(
x
−
x
0
)
+
B
(
y
−
y
0
)
+
C
(
z
−
z
0
)
=
0
, which we could write as Ax+By+Cz+D=0
A
x
+
B
y
+
C
z
+
D
=
0
, where D=−Ax0−By0−Cz0
D
=
−
A
x
0
−
B
y
0
−
C
z
0
.
This applet demonstrates the setup of the problem and the method we will use to derive a formula for the distance from the plane to the point P
P
.
Step-by-step explanation:
