prove that if two circles are tangent externally then the common internal tangent bisects a common external tangent.
1 answer:
Explanation:
Consider the attached diagram.
Circles A and B have mutual tangents CF and DE that intersect at point F.
We know that the two tangents to a circle from an external point are congruent, so ...
FE≅ FC
FD ≅ FC
By the transitive property of congruence, both segments congruent to FC are congruent to each other:
FD ≅ FE
Therefore, CF is a bisector of DE.
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So we start by combining like terms;
Now we have;
We then subtract 1 from both sides;
Then we divide both sides by 5;
Leaving the final answer to be h = -2
A. First let us start with the linear model.
The equation is in the form of y = m x + b
Calculating for the slope m:
m = (309,587 – 50,000) / (2007 – 1997)
m = 25,958.7
Subsituting:
y = 25,958.7 x + b
Taking x = 1997, y = 50,000. Solve for b:
50,000 = 25,958.7 (1997) + b
b = -51,789,523.9
The complete equation is therefore:
y = 25,958.7 x - 51,789,523.9
B. The exponential model has the following form:
y = a b^x
where a and b are constants
Taking x1= 1997, y1 = 50,000; x2 = 2007, y2 = 309,587
50,000 = a b^1997
309,587 = a b^2007
Combining in terms of a:
50,000 / b^1997 = 309,587 / b^2007
b^2007 / b^1997 = 309,587 / 50,000
b^10 = 6.19174
b = 1.2
Substituting:
y = a 1.2^x
Solving for a:
50,000 = a 1.2^1997
a = 3.75 x 10^-154
The complete equation is:
y = 3.75 x 10^-154 * 1.2^x
