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Key Stone Problem… Key Stone Problem… next Set 7 Part 1 © 2007 Herbert I. Gross

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You will soon be assigned five problems to test whether you have internalized the material in Lesson 7 part 1 of our algebra course. The Keystone Illustration below is a prototype of the problems you'll be doing. Work out the problem on your own. Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that could be used to solve the problem. Instructions for the Keystone Problem next © 2007 Herbert I. Gross

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As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

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next © 2007 Herbert I. Gross Note next By place value notation we mean, for example, writing 1,000 rather than 10 3. Write each of the following in place value notation.

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next Write in place value notation. (a) 2 × (10 3 ) 2 Keystone Illustration for Lesson 7 Part 1 Answer: 2,000,000 © 2007 Herbert I. Gross next

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Solution for Part a: By PEMDAS we work with the exponents before we multiply. That is, we read the problem as 2 × [(10 3 ) 2 ] and work within the parentheses first. next © 2007 Herbert I. Gross So we begin by using the rule (b m ) n = b mn ( b m ) n = b m n b = 10 10 m = 3 3 n = 22 next 10 3 2 ( ) 6 = 1,000,000

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Solution for Part a: If we now replace (10 3 ) 2 by 1,000,000 we see that it becomes… next © 2007 Herbert I. Gross 2 × (10 3 ) 2 = 1,000,000 next 2,000,000

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next © 2007 Herbert I. Gross Note We don't have to memorize the rules for exponents. That is, once we know the definitions involved with using bases and powers, we can always return to “basics”.

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next © 2007 Herbert I. Gross Note For example, in this problem, starting with (10 3 ) 2 and knowing that (b) 2 means (b) × (b), we see that (10 3 ) 2 means… 10 3 × 10 3 10 × 10 × 10 × and since 10 3 equals 10 × 10 × 10, we rewrite it as… next

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© 2007 Herbert I. Gross Note Because we can multiply in any order that we wish, we may eliminate the parentheses and write… ( 10 × 10 × 10 ) × ( 10 × 10 × 10 ) next 10 1 By definition the product of six factors of 10 is written as 10 6. In this way we have shown that (10 3 ) 2 = 10 6 10 2 10 3 10 4 10 5 10 6 next

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© 2007 Herbert I. Gross b m × b n = b m + n next Note to show that if b = 10 We could also have used the property that… 101010 m = 3 3 3 n = 3 3 3 which equals 10 6. = 10 6 next

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Write in place value notation. (b) (2 ×10 3 ) 2 Keystone Illustration for Lesson 7 Part 1 Answer: 4,000,000 © 2007 Herbert I. Gross next

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Solution for Part b: Since everything within a set of parentheses is considered to be one number, it is 2 × 10 3 that is being multiplied by itself. next © 2007 Herbert I. Gross Because we can multiply in any order that we wish, we may rewrite it as… (2×)(2×)×10 3 3 42 × 2 = 10 6 and 10 3 × 10 3 = next

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Solution for Part b: Hence… next © 2007 Herbert I. Gross (2 × 10 3 ) 2 = 4 × 10 6 = 4 × 1,000,000 = 4,000,000

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next © 2007 Herbert I. Gross Note Notice that parts (a) and (b) look alike except for the grouping symbols. When we write this 2 × (10 3 ) 2, our agreement tells us that we first raise 10 to the 3rd power and then multiply by 2. However, in writing this (2 ×10 3 ) 2, our agreement tells us first to multiply 10 3 by 2 and then raise the result to the 2nd power.

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next © 2007 Herbert I. Gross Note As usual there is more than one way to solve a problem. For example in the present problem, if we start within the parentheses first we can rewrite 2 × 10 3 as 2 × 1,000 or 2,000; and if we then square 2,000 we obtain 4,000,000

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next Write in place value notation. (c) (2 ×10 3 ) 2 × 5 × 10 6 Keystone Illustration for Lesson 7 Part 1 Answer: 20,000,000,000,000 © 2007 Herbert I. Gross next

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Solution for Part c: In part (b) we saw that (2 ×10 3 ) 2 = 4 × 10 6. So if we replace (2 × 10 3 ) 2 by 4 × 10 6 in the expression… next © 2007 Herbert I. Gross We obtain the equivalent expression… next (2 × 10 3 ) 2 × 5 × 10 6 4 × 10 6

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Solution for Part c: And since we can multiply in any order that we wish, we may rewrite the expression in the equivalent form… next © 2007 Herbert I. Gross Hence we obtain… (4×)(5×)×10 6 6 204 × 5 = 10 12 and 10 6 × 10 6 = next

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Solution for Part c: Every time we multiply a whole number by 10, we annex a zero. Hence to multiply a number by 10 12 we annex 12 zeroes, and we obtain... next © 2007 Herbert I. Gross next 20 × 10 12 = 20 000 000 000 0 00,,,,

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next © 2007 Herbert I. Gross Note Make sure you pay attention to the grouping symbols. For example in looking at (2 × 10 3 ) 2 × 5 × 10 it might be tempting to multiply the 2 inside the parentheses by the 5 outside of the parentheses. However, because the 2 is inside the parentheses, it is part of what's being squared. The 5 is multiplying 2 2, not 2.

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next © 2007 Herbert I. Gross Note Again notice that if you have not internalized the rules of arithmetic for exponents, you can still return to the basic definition. For example (2 ×10 3 ) × 5× 10 6 ( 2,000 ) 22 = 4,000,000 × 5 × 1,000,000 next It is now in the equivalent form.

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© 2007 Herbert I. Gross Note Finally, we multiply 4 by 5 to obtain 20 and then annex twelve 0's so that the expression may be rewritten as 20,000,000,000,000. × 5 × 1,000,000420,000,000 next

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Write in place value notation. (d) (2 ×10 3 ) 2 ÷ (2 × 10 4 ) Keystone Illustration for Lesson 7 Part 1 Answer: 200 © 2007 Herbert I. Gross next

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Solution for Part d: We already know that (2 × 10 3 ) 2 = 4,000,000. 2 × 10 4 means that we multiply 2 by four factors of 10; and since each time we multiply by 10 we annex a zero, we may rewrite 2 × 10 4 as 20,000. next © 2007 Herbert I. Gross Therefore we may rewrite the equation as… next 4,000,000 ÷ 20,000(2 × 10 3 ) 2 ÷ (2 × 10 4 )

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Solution for Part d: And if we now divide both numbers by 10,000 we obtain 400 ÷ 2 or 200. next © 2007 Herbert I. Gross next 200 400 ÷ 2 4,000,000 ÷ 20,000 next

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© 2007 Herbert I. Gross Note If the second set of parentheses in this problem had been omitted the problem would have been to rewrite (2 ×10 3 ) 2 ÷ 2 × 10 4 in place value notation. In this case the place value of the expression would have been… 4,0000,000 ÷ 2 × 10,000

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next © 2007 Herbert I. Gross Note In this case, since only multiplication and division are the only operations, the PEMDAS agreement tells us to read (2 ×10 3 ) 2 ÷ 2 × 10 4 from left to right. 2,000,000 × 10,000 4,000,000 ÷ 2 × 10,000 Accordingly, we would first divide 4,000,000 by 2 to obtain 2,000,000, and we then multiply 2,000,000 by 10,000 to obtain… = 20,000,000,000 next

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