Answer:
Using the concept of speed-distance-time, we find that the diameter of the circular path is 12 inches.
Step-by-step explanation:
Given that the object completes one revolution in 8 seconds. So, the circular speed of the object is 2πr/8 = πr/4 inches per seconds, where r is the radius of the circular path. The speed can be written as:
πr/4 inches per seconds = 60πr/4 inches per minute = 15πr inches per minute.
But, it is given that the circular speed of the object is 90π inches per minute. So,
90π = 15πr
r = 6 inches
Since, diameter is the double of radius, d = 2r = 12 inches.
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First, an introduction: If the equation of the circle were x^2 + y^2 = 144, then the center would be at (0,0) and the radius would be 12. Note that the distance from the center to P(10,10) is 10sqrt(2), or 14.14. Thus, in this example, P would be OUTSIDE the circle (since 14.14 is greater than the radius 12).
Now let's focus on <span>(x-1)^2 + (y-2)^2 =144. Let x = 12 as an example; find the corresponding y: 9^2 + (y-2)^2 = 144, or (y-2)^2 = 63, and so y-2 is approx. -8 or +8. Then y (for x = 12) is either approx. -10 or 6: (12,-10) or (12,6). Are these inside the circle or outside?
A better way to address this would be as follows:
Find the distance from the center (1, 2) to the point P(10,10). If this distance is less than 12, the point P is inside C; if greater than 12, P is outside C.
This distance is sqrt( (10-2)^2 + (10-1)^2 ), or sqrt (64+81) = sqrt(145).
This is LARGER than sqrt(144). Thus, P is OUTSIDE the circle C.</span>
i) The given function is
The factored form is
The domain are the values of x for which the function is defined.
ii) To find the vertical asymptotes, equate the denominator to zero.
iii) To find the roots, equate the numerator to zero.
The root is 
iv) To find the y-intercept, put 
into the function.
The y-intercept is 
v) The horizontal asymptote is given by;
The horizontal asymptote is 
vi) The function is not reducible. There are no holes.
vii) The given function is a proper rational function.
Proper rational functions do not have oblique asymptotes.
idk man im trying to get points so ty
An inverse variation can be represented by the equation
We have
Answer: The relationship does not show an inverse variation.
