Pre-Algebra Student Edition
Pre-Algebra Student Edition
1st Edition
ISBN: 9780078957734
Author: Glencoe/McGraw-Hill
Publisher: MCG
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Chapter 2.4, Problem 42PPS

a.

To determine

To calculate: The complete table that shows the altitude of the balloon every 5 minutes.

a.

Expert Solution
Check Mark

Answer to Problem 42PPS

The altitude of balloon every 5 minutes is tabulated below,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+6006005−15(5)+60052510−15(10)+60045015−15(15)+600375

Explanation of Solution

Given information:

The table,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+60060051015

Formula used:

Product of two real numbers is obtained when they are multiplied together. That is two or more numbers are separated by multiplication signs (),× or ⋅ .

Calculation:

Consider the given table,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+60060051015

Recall that Product of two real numbers is obtained when they are multiplied together. That is two or more numbers are separated by multiplication signs (),× or ⋅ .

So, apply it to calculate the values in table as,

When x=5 , value of y is calculated as,

  y=−15(5)+600=−75+600=525

When x=10 , value of y is calculated as,

  y=−15(10)+600=−150+600=450

When x=15 , value of y is calculated as,

  y=−15(15)+600=−225+600=375

The complete table is,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+6006005−15(5)+60052510−15(10)+60045015−15(15)+600375

Thus, the altitude of balloon every 5 minutes is tabulated below,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+6006005−15(5)+60052510−15(10)+60045015−15(15)+600375

b.

To determine

To describe: What the input value of 0 represents.

b.

Expert Solution
Check Mark

Answer to Problem 42PPS

The input value of 0 represents that the balloon did not fly yet and is still at its starting point.

Explanation of Solution

Given information:

A hot air balloon is at an altitude of 600 feet. It begins descending at a constant rate of 15 feet per minute.

The altitude of balloon every 5 minutes is tabulated in the table as,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+6006005−15(5)+60052510−15(10)+60045015−15(15)+600375

From the table, it is observed that input value 0 gives output as 600 .

And it is given that the hot air balloon is at an altitude of 600 feet.

So, input value of 0 represents that the balloon has not started flying yet and is still at starting point. It will start descending at the rate of 15 feet per minute.

c.

To determine

To graph: The order pairs of the following table,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+6006005−15(5)+60052510−15(10)+60045015−15(15)+600375

c.

Expert Solution
Check Mark

Explanation of Solution

Given information:

The table,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+6006005−15(5)+60052510−15(10)+60045015−15(15)+600375

Graph:

The altitude of balloon every 5 minutes is tabulated in the table as,

  Input x(minutes)y=−15x+600Output y(altitude)0−15(0)+6006005−15(5)+60052510−15(10)+60045015−15(15)+600375

From the table it is observed that the ordered pairs so formed are (0,600),(5,525),(10,450) and (15,375) .

Plot the ordered pairs on the graph as,

  Pre-Algebra Student Edition, Chapter 2.4, Problem 42PPS

Interpretation:

From the graph it is observed that as the time increases, the altitude of balloon decreases at the constant rate of 15 feet per minute.

d.

To determine

To calculate: The time taken by balloon to reach the ground.

d.

Expert Solution
Check Mark

Answer to Problem 42PPS

The time taken by balloon to reach the ground is 40 minutes .

Explanation of Solution

Given information:

A hot air balloon is at an altitude of 600 feet. It begins descending at a constant rate of 15 feet per minute.

The expression that shows that altitude of the balloon every 5 minutes y=−15x+600 .

Formula used:

Product of two real numbers is obtained when they are multiplied together. That is two or more numbers are separated by multiplication signs (),× or ⋅ .

Calculation:

Consider the given information A hot air balloon is at an altitude of 600 feet. It begins descending at a constant rate of 15 feet per minute.

The expression y=−15x+600 .

The time taken by balloon to reach ground will be calculated when y=0.

Recall that Product of two real numbers is obtained when they are multiplied together. That is two or more numbers are separated by multiplication signs (),× or ⋅ .

So, apply it to calculate the time taken by balloon to reach ground as,

  y=−15x+6000=−15x+60015x=600x=40

Thus, the time taken by balloon to reach ground is 40 minutes .

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Chapter 2.4, Problem 41PPS
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Chapter 2.4, Problem 43HP

Chapter 2 Solutions

Pre-Algebra Student Edition

Ch. 2.1 - Prob. 1AGPCh. 2.1 - Prob. 1BGPCh. 2.1 - Prob. 2AGPCh. 2.1 - Prob. 2BGPCh. 2.1 - Prob. 3GPCh. 2.1 - Prob. 4AGPCh. 2.1 - Prob. 4BGPCh. 2.1 - Prob. 5GPCh. 2.1 - Prob. 1CYUCh. 2.1 - Prob. 2CYU
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Prob. 26STPCh. 2.7 - Prob. 27SRCh. 2.7 - Prob. 28SRCh. 2.7 - Prob. 29SRCh. 2.7 - Prob. 30SRCh. 2.7 - Prob. 31SRCh. 2.7 - Prob. 32SRCh. 2.7 - Prob. 33SRCh. 2.7 - Prob. 34SCh. 2.7 - Prob. 35SCh. 2.7 - Prob. 36SCh. 2.7 - Prob. 37SCh. 2 - Prob. 1QCCh. 2 - Prob. 2QCCh. 2 - Prob. 3QCCh. 2 - Prob. 4QCCh. 2 - Prob. 5QCCh. 2 - Prob. 6QCCh. 2 - Prob. 7QCCh. 2 - Prob. 8QCCh. 2 - Prob. 9QCCh. 2 - Prob. 10QCCh. 2 - Prob. 11QCCh. 2 - Prob. 12QCCh. 2 - Prob. 13QCCh. 2 - Prob. 14QCCh. 2 - Prob. 15QCCh. 2 - Prob. 16QCCh. 2 - Prob. 1MCQCh. 2 - Prob. 2MCQCh. 2 - Prob. 3MCQCh. 2 - Prob. 4MCQCh. 2 - Prob. 5MCQCh. 2 - Prob. 6MCQCh. 2 - Prob. 7MCQCh. 2 - Prob. 8MCQCh. 2 - Prob. 9MCQCh. 2 - Prob. 10MCQCh. 2 - Prob. 11MCQCh. 2 - Prob. 12MCQCh. 2 - Prob. 13MCQCh. 2 - Prob. 14MCQCh. 2 - Prob. 15MCQCh. 2 - Prob. 16MCQCh. 2 - Prob. 17MCQCh. 2 - Prob. 18MCQCh. 2 - Prob. 19MCQCh. 2 - Prob. 20MCQCh. 2 - Prob. 21MCQCh. 2 - Prob. 22MCQCh. 2 - Prob. 23MCQCh. 2 - Prob. 24MCQCh. 2 - Prob. 25MCQCh. 2 - Prob. 1SGRCh. 2 - Prob. 2SGRCh. 2 - Prob. 3SGRCh. 2 - Prob. 4SGRCh. 2 - Prob. 5SGRCh. 2 - Prob. 6SGRCh. 2 - Prob. 7SGRCh. 2 - Prob. 8SGRCh. 2 - Prob. 9SGRCh. 2 - Prob. 10SGRCh. 2 - Prob. 11SGRCh. 2 - Prob. 12SGRCh. 2 - Prob. 13SGRCh. 2 - Prob. 14SGRCh. 2 - Prob. 15SGRCh. 2 - Prob. 16SGRCh. 2 - Prob. 17SGRCh. 2 - Prob. 18SGRCh. 2 - Prob. 19SGRCh. 2 - Prob. 20SGRCh. 2 - Prob. 21SGRCh. 2 - Prob. 22SGRCh. 2 - Prob. 23SGRCh. 2 - Prob. 24SGRCh. 2 - Prob. 25SGRCh. 2 - Prob. 26SGRCh. 2 - Prob. 27SGRCh. 2 - Prob. 28SGRCh. 2 - Prob. 29SGRCh. 2 - Prob. 30SGRCh. 2 - Prob. 31SGRCh. 2 - Prob. 32SGRCh. 2 - Prob. 33SGRCh. 2 - Prob. 34SGRCh. 2 - Prob. 35SGRCh. 2 - Prob. 36SGRCh. 2 - Prob. 37SGRCh. 2 - Prob. 38SGRCh. 2 - Prob. 39SGRCh. 2 - Prob. 40SGRCh. 2 - Prob. 41SGRCh. 2 - Prob. 42SGRCh. 2 - Prob. 43SGRCh. 2 - Prob. 44SGRCh. 2 - Prob. 45SGRCh. 2 - Prob. 46SGRCh. 2 - Prob. 47SGRCh. 2 - Prob. 48SGRCh. 2 - Prob. 49SGRCh. 2 - Prob. 50SGRCh. 2 - Prob. 51SGRCh. 2 - Prob. 52SGRCh. 2 - Prob. 53SGRCh. 2 - Prob. 54SGRCh. 2 - Prob. 55SGRCh. 2 - Prob. 56SGRCh. 2 - Prob. 57SGRCh. 2 - Prob. 58SGRCh. 2 - Prob. 59SGRCh. 2 - Prob. 60SGRCh. 2 - Prob. 61SGRCh. 2 - Prob. 1PTCh. 2 - Prob. 2PTCh. 2 - Prob. 3PTCh. 2 - Prob. 4PTCh. 2 - Prob. 5PTCh. 2 - Prob. 6PTCh. 2 - Prob. 7PTCh. 2 - Prob. 8PTCh. 2 - Prob. 9PTCh. 2 - Prob. 10PTCh. 2 - Prob. 11PTCh. 2 - Prob. 12PTCh. 2 - Prob. 13PTCh. 2 - Prob. 14PTCh. 2 - Prob. 15PTCh. 2 - Prob. 16PTCh. 2 - Prob. 17PTCh. 2 - Prob. 18PTCh. 2 - Prob. 19PTCh. 2 - Prob. 20PTCh. 2 - Prob. 21PTCh. 2 - Prob. 22PTCh. 2 - Prob. 23PTCh. 2 - Prob. 24PTCh. 2 - Prob. 25PTCh. 2 - Prob. 26PTCh. 2 - Prob. 27PTCh. 2 - Prob. 28PTCh. 2 - Prob. 29PTCh. 2 - Prob. 30PTCh. 2 - Prob. 31PTCh. 2 - Prob. 32PTCh. 2 - Prob. 33PTCh. 2 - Prob. 1ECh. 2 - Prob. 2ECh. 2 - Prob. 3ECh. 2 - Prob. 4ECh. 2 - Prob. 5ECh. 2 - Prob. 1STPCh. 2 - Prob. 2STPCh. 2 - Prob. 3STPCh. 2 - Prob. 4STPCh. 2 - Prob. 5STPCh. 2 - Prob. 6STPCh. 2 - Prob. 7STPCh. 2 - Prob. 8STPCh. 2 - Prob. 9STPCh. 2 - Prob. 10STPCh. 2 - Prob. 11STPCh. 2 - Prob. 12STPCh. 2 - Prob. 13STPCh. 2 - Prob. 14STP

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