1. In the example above, what is the height of the tunnel 12 ft away from the center towards the other direction?2. What is the height of the tunnel 20 ft away from the center? What happens to the height of the tunnel as you go away from the center?3. How far from the center of the tunnel have a height of exactly 16 ft? Before solving problems involving conic sections, it is a must that we know what algebraic operations can bedone with their equations. In the previous modules, we've done finding the discriminant, squaring, completingthe square, rewriting from one form to another, etc. One thing we haven't done yet is "evaluation".To evaluate these conic equations, we simply substitute their variables with constants, perform any operationand then simplify.Example 1.The height of a semi-circular twnnel is 24 ft at the center. What is itsheight 12 ft away from the center?It is obvious that this problem deals with a cirde, so we use itsequation (x – h)² + (y – k)? = r² to solve this problem. From theproblem, we know the radius r = 24 ft. Since the center (h, k) is unknown in the problem, we assume that itis located at (0, 0) and that h = 0, and k = 0. This makes the equation (x – 0)² + (y – 0)² = 24°. This can befurther simplified resulting in a shorter equation that is x² + y² = 576. Looking at the illustration, we canobserve that the part of the circle where x is at zero (x = 0), y is at 24 (y = 24). That means we have to findy when x is at 12. To do this, we evaluate the equation using 12 as a value of x.Solution:12, 20.78)x + y = 576122 + y = 576144 + y? - 576Y? = 576 – 144Y = 4321818141220.7810Y = /432Y = 20.78 When x = 12, y = 20.78 and 20.78 ftis the height of the tunnel 12 ft awayfrom the center.(12. 0)16 18 20This example is for circles but the same principle can be applied to other types.

From the example above, We can see that we are using the…

1. In the example above, what is the height of the tunnel 12 ft away from the center towards the other direction? 2. What is the height of the tunnel 20 ft away from the center? What happens to the height of the tunnel as you go away from the center? 3. How far from the center of the tunnel have a height of exactly 16 ft?

1. In the example above, what is the height of the tunnel 12 ft away from the center towards the other direction? 2. What is the height of the tunnel 20 ft away from the center? What happens to the height of the tunnel as you go away from the center? 3. How far from the center of the tunnel have a height of exactly 16 ft?

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

1. In the example above, what is the height of the tunnel 12 ft away from the center towards the other direction?
2. What is the height of the tunnel 20 ft away from the center? What happens to the height of the tunnel as you go away from the center?
3. How far from the center of the tunnel have a height of exactly 16 ft?

Before solving problems involving conic sections, it is a must that we know what algebraic operations can be
done with their equations. In the previous modules, we've done finding the discriminant, squaring, completing
the square, rewriting from one form to another, etc. One thing we haven't done yet is "evaluation".
To evaluate these conic equations, we simply substitute their variables with constants, perform any operation
and then simplify.
Example 1.
The height of a semi-circular twnnel is 24 ft at the center. What is its
height 12 ft away from the center?
It is obvious that this problem deals with a cirde, so we use its
equation (x – h)² + (y – k)? = r² to solve this problem. From the
problem, we know the radius r = 24 ft. Since the center (h, k) is unknown in the problem, we assume that it
is located at (0, 0) and that h = 0, and k = 0. This makes the equation (x – 0)² + (y – 0)² = 24°. This can be
further simplified resulting in a shorter equation that is x² + y² = 576. Looking at the illustration, we can
observe that the part of the circle where x is at zero (x = 0), y is at 24 (y = 24). That means we have to find
y when x is at 12. To do this, we evaluate the equation using 12 as a value of x.
Solution:
12, 20.78)
x + y = 576
122 + y = 576
144 + y? - 576
Y? = 576 – 144
Y = 432
<equation>
<substitution>
18
18
14
12
20.78
10
Y = /432
Y = 20.78 <final result>
When x = 12, y = 20.78 and 20.78 ft
is the height of the tunnel 12 ft away
from the center.
(12. 0)
16 18 20
This example is for circles but the same principle can be applied to other types.

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