Answer Equation of the safe working load for a rope is f ( x ) = − 180 x 2 . f ( 10 ) = 18000 . Explanation Given information: Every rope has a safe working load. A rope should not be used to lift a weight greater than its safe working load. The table shows the safe working loads S (in pounds) for ropes with circumference C (in inches ).: Calculation Because second differences are constant so we can modal the data with a quadratic function f ( x ) = a x 2 + b x + c . f ( 0 ) f ( 1 ) f ( 2 ) f ( 3 ) 0 180 720 1620 180 540 900 360 360 f ( x ) = a x 2 + b x + c ....... ( 1 ) . f ( 0 ) = a × 0 2 + b × 0 + c . 0 = 0 + c . c = 0 . f ( 1 ) = a × 1 2 + b × 1 = a + b . a + b = 180 ......... ( 2 ) f ( 2 ) = a × 2 2 + b × 2 = 4 a + 2 b . 4 a + 2 b = 180........ ( 3 ) solving equation (2) and (3) . a = 180 and b = 0 . putting a = 180 , b = 0 , c = 0 in equation ( 1 ) . f ( x ) = 180 x 2 . So, f ( 10 ) = 180 × 10 2 = 180 × 100 = 18000 .

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

1st Edition

ISBN: 9781680331165

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 2.4, Problem 23E

To determine

Write an equation for the safe working load for a rope. Find the safe working load for a rope that has a circumference of 10 inches.

Expert Solution & Answer

Answer to Problem 23E

Equation of the safe working load for a rope is f(x)=−180x2.

f(10)=18000 .

Explanation of Solution

Given information: Every rope has a safe working load. A rope should not be used to

lift a weight greater than its safe working load. The table shows the safe working loads S (in pounds) for ropes with circumference C (in inches).:

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 2.4, Problem 23E

Calculation

Because second differences are constant so we can modal the data with a quadratic function f(x)=ax2+bx+c .

f(0)		f(1)		f(2)		f(3)
0		180		720		1620
	180		540		900
		360		360

f(x)=ax2+bx+c .......(1).f(0)=a×02+b×0+c .0=0+c .c=0 .f(1)=a×12+b×1=a+b .a+b=180 .........(2)f(2)=a×22+b×2=4a+2b .4a+2b=180........(3)solving equation (2) and (3) .a=180 and b=0 .putting a=180 ,b=0 ,c=0 in equation (1) .f(x)=180x2 .

So,

f(10)=180×102=180×100=18000 .

Chapter 2.4, Problem 22E

Chapter 2.4, Problem 24E

Chapter 2 Solutions

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

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