<h3>Refer to the diagram below</h3>
- Draw one smaller circle inside another larger circle. Make sure the circle's edges do not touch in any way. Based on this diagram, you can see that any tangent of the smaller circle cannot possibly intersect the larger circle at exactly one location (hence that inner circle tangent cannot be a tangent to the larger circle). So that's why there are no common tangents in this situation.
- Start with the drawing made in problem 1. Move the smaller circle so that it's now touching the larger circle at exactly one point. Make sure the smaller circle is completely inside the larger one. They both share a common point of tangency and therefore share a common single tangent line.
- Start with the drawing made for problem 2. Move the smaller circle so that it's partially outside the larger circle. This will allow for two different common tangents to form.
- Start with the drawing made for problem 3. Move the smaller circle so that it's completely outside the larger circle, but have the circles touch at exactly one point. This will allow for an internal common tangent plus two extra external common tangents.
- Pull the two circles completely apart. Make sure they don't touch at all. This will allow us to have four different common tangents. Two of those tangents are internal, while the others are external. An internal tangent cuts through the line that directly connects the centers of the circles.
Refer to the diagram below for examples of what I mean.
set them equal to their sum. which in this case is 141 then solve for x and plug it into your original equation to get your answer.
1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :
i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.
ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:
a^{2}=b ^{2}+c ^{2}-2bc(cosA)
2.
20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA) 400=81+169-234(cosA) 150=-234(cosA) cosA=150/-234= -0.641
3. m(A) = Arccos(-0.641)≈130°,
4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc
The linear equation that models the cost for having x additional family members added is:
c(x) = $75 + $10.99*x
<h3>
How to get the linear equation?</h3>
Here we know that the plan has a fixed price of $75 plus $10.99 for each family member added beyond the primary account holder.
Then if there are x family members added, the cost will be:
$75 + $10.99*x
Then the linear equation that models the cost for having x additional family members added is:
c(x) = $75 + $10.99*x
If you want to learn more about linear equations:
brainly.com/question/1884491
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Answer:
-1
Step-by-step explanation:
slope = y2-y1/x2-x1
(5,5) (7,3)
slope = 5-3/5-7 = 2/-2 = -1