Answer:
Two distinct concentric circles: 0 max solutions
Two distinct parabolas: 4 max solutions
A line and a circle: 2 max solution
A parabola and a circle: 4 max solutions
Step-by-step explanation:
<u>Two distinct concentric circles:</u>
The maximum number of solutions (intersections points) 2 distinct circles can have is 0. You can see an example of it in the first picture attached. The two circles are shown side to side for clarity, but when they will be concentric, they will have same center and they will be superimposed. So there can be ZERO max solutions for that.
<u>Two distinct parabolas:</u>
The maximum solutions (intersection points) 2 distinct parabolas can have is 4. This is shown in the second picture attached. <em>This occurs when two parabolas and in perpendicular orientation to each other. </em>
<u>A line and a circle:</u>
The maximum solutions (intersection points) a line and a circle can have is 2. See an example in the third picture attached.
<u>A parabola and a circle:</u>
The maximum solutions (intersection points) a parabola and a circle can have is 4. If the parabola is <em>compressed enough than the diameter of the circle</em>, there can be max 4 intersection points. See the fourth picture attached as an example.
Answer:
The Correct Statement is
• Use the distance formula to prove the lengths of the opposite sides are the same.
Step-by-step explanation:
The statement explains how you could use coordinate geometry to prove the opposite sides of a quadrilateral are congruent is,
• Use the distance formula to prove the lengths of the opposite sides are the same.
Distance Formula is given by
For Parallel Lines:
Use the slope formula to prove the slopes of the opposite sides are the same.
For Perpendicular Lines:
Use the slope formula to prove the slopes of the opposite sides are opposite reciprocals.
*See the figure in the attachment below
Answer:
346.4 ft
Step-by-step Explanation:
The distance between the boats, CD = BD - BC
Using trigonometric ratio formula, find the length of BD in right triangle ∆ABD, and also find the length of BC in right triangle ABC.
Length of BD in triangle ABD:
Opposite length to <30° = 300 ft
Adjacent length = BD
Divide both sides by tan(30)
(nearest tenth)
Length of BC in triangle ABC:
Opposite length to <60° = 300 ft
Adjacent length = BC
Divide both sides by tan(60)
(nearest tenth)
distance between, CD, between the boats = BD - BC
= 519.6 - 173.2 = 346.4 ft
((4a+4)/(2a))((a^2)/(a+1))
(2(2a+2)/2(a))((a^2)/(a+1))
((2a+2)(a^2))/(a(a+1))
(2(a+1)(a^2))/(a(a+1))
(2a^2)/a
=2a
Answer:
a) 20 ways
b) 10 ways
Step-by-step explanation:
When the order of selection/choice matters, we use Permutations to find the number of ways and if the order of selection/choice does not matter, we use Combinations to find the number of ways.
Part a)
We have to chose 2 objects from a group of 5 objects and order of choice matters. This is a problem of permutations, so we have to find 5P2
General formula of permutations of n objects taken r at time is:
Using the value of n=5 and r=2, we get:
Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice matters.
Part b)
Order of choice does not matter in this case, so we will use combinations to find the number of ways of choosing 2 objects from a group of 5 objects which is represented by 5C2.
The general formula of combinations of n objects taken r at a time is:
Using the value of n=5 and r=2, we get:
Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice does not matters.
