Fireworks. Two firework shells are fired upward at the same time from different platforms. The height, after t seconds, of the first shell is (-16t² + 160t + 3) feet. The height, after t seconds, of a higher-flying second shell is (-16t2 + 200t + 1) feet. (a) Find a polynomial that represents the difference in the heights of the shells. (b) In 5 seconds, the first shell reaches its peak and explodes. How much higher is the second shell at that time? To find the difference in their heights, we will subtract the height of the first shell from the height of the higher-flying second shell. The key word difference indicates that we should subtract the polynomials. (a) Since the height of the higher flying second shell is represented by -16t2 + 200t + 1 and the height of the lower flying shell is represented by –16t2 + 160t + 3, we can fine their difference by performing the following subtraction. (-16t2 + 200t + 1) – (-16t2 + 160t + 3) = -16t2 + 200t + 1 + 16t2 – 160t – Change the sign of each term of –16t + 160t + 3 and remove parentheses. t - 2 Combine like terms. The difference in the heights of the shellst seconds after being fired is (40t – 2) feet. (b) To find the difference in their heights after 5 seconds, we will evaluate the polynomial found in part (a) at a value of 5 seconds. If we substitute 5 for t, we have 40t – 2 = 40( ) - 2 = 200 – 2 =[ When the first shell explodes, the second shell will be 198 feet higher than the first shell.
Fireworks. Two firework shells are fired upward at the same time from different platforms. The height, after t seconds, of the first shell is (-16t² + 160t + 3) feet. The height, after t seconds, of a higher-flying second shell is (-16t2 + 200t + 1) feet. (a) Find a polynomial that represents the difference in the heights of the shells. (b) In 5 seconds, the first shell reaches its peak and explodes. How much higher is the second shell at that time? To find the difference in their heights, we will subtract the height of the first shell from the height of the higher-flying second shell. The key word difference indicates that we should subtract the polynomials. (a) Since the height of the higher flying second shell is represented by -16t2 + 200t + 1 and the height of the lower flying shell is represented by –16t2 + 160t + 3, we can fine their difference by performing the following subtraction. (-16t2 + 200t + 1) – (-16t2 + 160t + 3) = -16t2 + 200t + 1 + 16t2 – 160t – Change the sign of each term of –16t + 160t + 3 and remove parentheses. t - 2 Combine like terms. The difference in the heights of the shellst seconds after being fired is (40t – 2) feet. (b) To find the difference in their heights after 5 seconds, we will evaluate the polynomial found in part (a) at a value of 5 seconds. If we substitute 5 for t, we have 40t – 2 = 40( ) - 2 = 200 – 2 =[ When the first shell explodes, the second shell will be 198 feet higher than the first shell.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter8: Polynomials
Section8.1: Adding And Subtracting Polynomials
Problem 59PPS
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Transcribed Image Text:Fireworks. Two firework shells are fired upward at the same time from different platforms. The height, after t seconds, of the first shell is (-16t2 + 160t + 3) feet. The height,
after t seconds, of a higher-flying second shell is (-16t + 200t + 1) feet.
(a) Find a polynomial that represents the difference in the heights of the shells.
(b) In 5 seconds, the first shell reaches its peak and explodes. How much higher is the second shell at that time?
To find the difference in their heights, we will subtract the height of the first shell from the height of the higher-flying second shell.
The key word difference indicates that we should subtract the polynomials.
(a) Since the height of the higher flying second shell is represented by -16t2 + 200t + 1 and the height of the lower flying shell is represented by -16t2 + 160t + 3, we can find
their difference by performing the following subtraction.
(-16t2 + 200t + 1) – (-16t2 + 160t + 3)
= -16t2 + 200t + 1 + 16t?
160t
Change the sign of each term of –16t2 + 160t + 3 and remove parentheses.
t -
2
Combine like terms.
The difference in the heights of the shellst seconds after being fired is (40t – 2) feet.
(b) To find the difference in their heights after 5 seconds, we will evaluate the polynomial found in part (a) at a value of 5 seconds. If we substitute 5 for t, we have
40t – 2 = 40( ) - 2 = 200 – 2 =
When the first shell explodes, the second shell will be 198 feet higher than the first shellI.
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