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High School Math Algebra BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

The time for which the fish has to swim away.

The time for which the fish has to swim away.

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

15th Edition

ISBN: 9781608408405

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Cengage Learning

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Question

Chapter 3.4, Problem 67E

a.

To determine

To calculate: The time for which the fish has to swim away.

a.

Expert Solution

Answer to Problem 67E

The required time for which the fish has to swim away is 0.97 seconds.

Explanation of Solution

Given information:

The initial height, h0=100 feet

The initial vertical velocity, v0=−88 feet per second

Formula used:

Quadratic formula:

x=−b±b2−4ac2a

Calculation:

The initial height is h0=100 feet.

The initial vertical velocity is v0=−88 feet per second.

To find the required time, use the equation h=−16t2+v0t+h0 .

Substitute −88 for v0 , 100 for h0 and 0 for h in the above equation.

0=−16t2−88t+100−16t24−88t4+1004=04−4t2−22t+25=04t2+22t−25=0

To solve the above equation for t , use the formula x=−b±b2−4ac2a .

t=−22±222−4(4)(−25)2(4)=−22±484+4008=−22±8848=−22±22218

Solve for t .

t=−22±22218t=−11±2214

Now, t=−11+2214=−11+14.864=0.965≈0.97

And, t=−11−2214=−11−14.864=−6.465

But time cannot be negative.

Hence, the required time for which the fish has to swim away is 0.97 seconds.

b.

To determine

To determine: The bird which will reach the fish first.

b.

Expert Solution

Answer to Problem 67E

The first bird will reach the fish first.

Explanation of Solution

Given information:

The initial height, h0=84 feet

The initial vertical velocity, v0=−70 feet per second

Formula used:

Quadratic formula:

x=−b±b2−4ac2a

Calculation:

The initial height is h0=84 feet.

The initial vertical velocity is v0=−70 feet per second.

To find the required time, use the equation h=−16t2+v0t+h0 .

Substitute −70 for v0 , 84 for h0 and 0 for h in the above equation.

0=−16t2−70t+8416t22+70t2−842=028t2+35t−42=0

To solve the above equation for t , use the formula x=−b±b2−4ac2a .

t=−35±352−4(8)(−42)2(8)=−35±1225+134416=−35±256916

Solve for t .

t=−35+256916t=−35+50.6816t=15.6816t=0.98

And, t=−35−256916t=−35−50.6816t=−85.6816t=−5.355

But time cannot be negative.

So, the time is 0.98 seconds.

Hence, the first bird will reach the fish first.

Chapter 3.4, Problem 66E

Chapter 3.4, Problem 68E

Chapter 3 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 39E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 41E Ch. 3.1 - Prob. 42E Ch. 3.1 - Prob. 43E Ch. 3.1 - Prob. 44E Ch. 3.1 - Prob. 45E Ch. 3.1 - Prob. 46E Ch. 3.1 - Prob. 47E Ch. 3.1 - Prob. 48E Ch. 3.1 - Prob. 49E Ch. 3.1 - Prob. 50E Ch. 3.1 - Prob. 51E Ch. 3.1 - Prob. 52E Ch. 3.1 - Prob. 53E Ch. 3.1 - Prob. 54E Ch. 3.1 - Prob. 55E Ch. 3.1 - Prob. 56E Ch. 3.1 - Prob. 57E Ch. 3.1 - Prob. 58E Ch. 3.1 - Prob. 59E Ch. 3.1 - Prob. 60E Ch. 3.1 - 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