Let the radius of the circle be r. Then the line from the external point through the center of the circle which extends to the far point on the circle has length 3r .By the tangent - secant theorem
t^2 = 3r * r = 3r^2 ( where t is the length of the tangent).
So t = √(3r^2) = √3r answer.
Answer: The equation is y = -6*x
Step-by-step explanation:
I suppose that we want to find the equation for a line that passes through the point (-1, 6) and the origin (remember that the origin is the point (0,0))
A general linear equation is written as:
y = a*x + b
Where a is the slope and b is the y-intercept.
If this line passes through the points (x₁, y₁) and (x₂, y₂), then the slope of the line is equal to:
a = (y₂ - y₁)/(x₂ - x₁)
Now we know that our line passes through the points (0, 0) and (-1, 6), then the slope is:
a = (6 - 0)/(-1 - 0) = 6/-1 = -6
Then our equation is something like:
y = -6*x + b
To find the value of b we can use the fact that this line passes through the point (0, 0).
This means that when x = 0, y is also equal to zero.
If we replace these values in the equation we get:
0 = -6*0 + b
0 = b
Then our equation is:
y = -6*x
In the default window of a graphing calculator, there is only one intersection that you see.
However, if you zoom out, you will see that they are 3 intersections to the pair of equations.
Answer:
D:3
Step-by-step explanation:
Answer:
D. 12/27
Step-by-step explanation:
Ratios are just like fractions
12 ÷ 3 = 4
27 ÷ 3 = 9
4/9 is 12/27 simplified by 3
