Given:
The equation of a circle is
A tangent line l to the circle touches the circle at point P(12,5).
To find:
The gradient of the line l.
Solution:
Slope formula: If a line passes through two points, then the slope of the line is
Endpoints of the radius are O(0,0) and P(12,5). So, the slope of radius is
We know that, the radius of a circle is always perpendicular to the tangent at the point of tangency.
Product of slopes of two perpendicular lines is always -1.
Let the slope of tangent line l is m. Then, the product of slopes of line l and radius is -1.
Therefore, the gradient or slope of the tangent line l is 
.
Answer:
the answer is d because it's reflects
The given equations is: a=b+4
Given the following clues:
a=24+4 --- 28
a=32+4 --- 36
a= 40+4 --- 44
Answer:
t=0.64
Step-by-step explanation:
h = -16t^2 +4t +4
We want h =0 since it is hitting the ground
0 = -16t^2 +4t +4
Using the quadratic formula
a = -16 b = 4 c=4
-b ± sqrt( b^2 -4ac)
----------------------------
2a
-4 ± sqrt( 4^2 -4(-16)4)
----------------------------
2(-16)
-4 ± sqrt( 16+ 256)
----------------------------
-32
-4 ± sqrt( 272)
----------------------------
-32
-4 ± sqrt( 16*17)
----------------------------
-32
-4 ± sqrt( 16) sqrt(17)
----------------------------
-32
-4 ± 4 sqrt(17)
----------------------------
-32
Divide by -4
1 ± sqrt(17)
----------------------------
8
To the nearest hundredth
t=-0.39
t=0.64
Since time cannot be negative
t=0.64
