## Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 2 - Solving Equations - 2-10 Change Expressed as a Percent - Practice and Problem-Solving Exercises - Page 150: 41

#### Answer

4a. 21% 4b. 21% 4c. 21%

#### Work Step by Step

4a. $A=(4)(4)=16$ $4\times10%=4$$\times0.1=0.4 Add 0.4 to the already given length to the new length of 4.4 A=(4.4)(4.4)=19.36 This is the new area p%=\frac{19.36-16}{16} =\frac{3.36}{16} 0.21 or 21% 4b. A=(6)(6)=36 6\times10%$$=6\times0.1=0.6$ Add 0.6 to the old length to get the new length $6+0.6=6.6$ $A=(6.6)(6.6)=43.56$ This is the new area of the square percent increase new amount-original amount/original amount p%=$\frac{43.56-36}{36}$ $=\frac{7.56}{36}$ $0.21$ or $21$% 4c. 21% The new length of 10 is 1.1 times as great as the original length $10\times10$%=0.1 $10+0.1=10.1$ $\frac{10.1}{10}=1.1$ $1.1^2=1.21$ or 121% This is a 21% increase over the original amount of 100%

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