a.
the maximum image width that can be projected.
a.

Answer to Problem 29PPE
y = 1. 8x
Explanation of Solution
Given Information
A projector and a screen, of which the distance between is ’12 feet’ and the width of the image described as a condition that for obtaining the required distance we have to multiply the image width by 1.8.
Calculation
Let, the distance of projector to the screen be ‘ y’ and, the desired width of image be ‘ x’ .
Now, we will establish a relation between distance and width of image as a function of width.
Since, the manual clearly states that for requiring distance, we have to multiply the desired width by 1.8
So,
b.
that can the projector able to project the image
b.

Answer to Problem 29PPE
not possible
Explanation of Solution
Given Information
A projector and a screen, of which the distance between is ’12 feet’ and the width of the image described as a condition that for obtaining the required distance we have to multiply the image width by 1.8.
Calculation:
For an image of 7 feet, we have to put the value of x = 7 feet in our established equation,
⇒
So, the distance comes out to be 12.6 feet.
But the maximum Length of hall is 12 feet.
Conclusion
So, it is not possible to project an image of 7 feet.
c.
that can the projector able to project the image
c.

Answer to Problem 29PPE
6.6 feet
Explanation of Solution
Given Information
A projector and a screen, of which the distance between is ’12 feet’ and the width of the image described as a condition that for obtaining the required distance we have to multiply the image width by 1.8.
Calculation:
The maximum image width, which can be projected in room, shall be calculated as
⇒
We are given,
So,
Conclusion So, the maximum image width that can be obtained is 6.6 feet.
Chapter 4 Solutions
High School Math 2015 Common Core Algebra 1 Student Edition Grade 8/9
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