INTERMEDIATE ALGEBRA ALEKS 360 ACCESS >
INTERMEDIATE ALGEBRA ALEKS 360 ACCESS >
5th Edition
ISBN: 9781260680805
Author: Miller
Publisher: MCG
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Chapter R3, Problem 1PE
To determine

The correct word or phrase to fill in the blank.

Expert Solution & Answer
Check Mark

Answer to Problem 1PE

a. Opposites

b. |a|, 0

c. Base, n

d. Radical, square

e. Negative, positive

f. Positive, negative

g. (b), positive

h. 1a, 1

i. 0

Explanation of Solution

a.

The opposite numbers are the numbers which differ in sign and are at an equal distance from 0. Therefore, the given statement can be completed as “Two numbers that are the same distance from 0 but on opposite sides of 0 on the number line, are called opposites.”

b.

The absolute value of a real number is denoted by writing the number within two vertical bars (“|” mark) and it is the distance between 0 and that number. So, the given statement can be completed as “The absolute value of a real number, a, is denoted by |a| and is the distance between a and 0on the number line.”

c.

Exponential numbers are written as xa , where x is the base and a is the exponent. For example, in 25 , 2 is the base and 5 is the exponent.

Therefore, the given statement can be completed as “Given the expression bn , the value b is called the base and n is called the exponent or power.”

d.

The square root of a number x is represented by x , where the symbol is called a radical.

Therefore, the given incomplete statement can be completed as “The symbol is called a radicalsign and is used to find the principle squareroot of a nonnegative real number.”

e.

The sum of two negative numbers is negative and product of two negative numbers is a positive number. So, the statement can be completed as “If a and b are both negative, then a+b will be negativeand ab will be positive.”

f.

The addition of two numbers of opposite signs are either positive or negative depending on the sign of the number with higher absolute value. For example, the addition of 2 and 6 will give 4 but addition of 2 and 6 will give 4. So, the given statement can be completed as shown below.

If a<0andb>0,andif|a|<|b| , then the sign of a+b will be positive and the sign of ab will be negative.

g.

Subtracting a number from another number is same as adding the opposite of the first number. And, addition of two numbers of opposite signs are either positive or negative depending on the sign of the number with higher absolute value.

The expression ab=a+(b)_. If a>0andb<0 , then the sign of ab is positive.

h.

The reciprocal numbers are the numbers whose product is equal to 1. And the reciprocal of a given number can be written by the number in the denominator over 1.

So, the statement can be completed as “If a is a nonzero real number, then the reciprocal of a is 1a_. The product of a number and its reciprocal is 1_ .”

i.

Any number multiplied to 0 gives the result 0. So, in the product ab of two numbers a and b, if any of the two numbers is zero, the product will be 0. So, the statement can be completed as “If either a or b is zero then ab=0_ .”

j.

If 0 is divided by any nonzero number the quotient is always 0. But, if a nonzero number is divided by 0, the quotient is undefined because the division by 0 is not defined. So, the statement can be completed as “If a=0andb0 , then ab=0_ and ba is undefined.”

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Chapter R3 Solutions

INTERMEDIATE ALGEBRA ALEKS 360 ACCESS >

Ch. R3 - Prob. 11PECh. R3 - Prob. 12PECh. R3 - Prob. 13PECh. R3 - Prob. 14PECh. R3 - Prob. 15PECh. R3 - Prob. 16PECh. R3 - Prob. 17PECh. R3 - Prob. 18PECh. R3 - Prob. 19PECh. R3 - Prob. 20PECh. R3 - Prob. 21PECh. R3 - Prob. 22PECh. R3 - Prob. 23PECh. R3 - Prob. 24PECh. R3 - Prob. 25PECh. R3 - Prob. 26PECh. R3 - Prob. 27PECh. R3 - Prob. 28PECh. R3 - Prob. 29PECh. R3 - Prob. 30PECh. R3 - Prob. 31PECh. R3 - Prob. 32PECh. R3 - Prob. 33PECh. R3 - Prob. 34PECh. R3 - Prob. 35PECh. R3 - Prob. 36PECh. R3 - Prob. 37PECh. R3 - Prob. 38PECh. R3 - Prob. 39PECh. R3 - Prob. 40PECh. R3 - Prob. 41PECh. R3 - Prob. 42PECh. R3 - Prob. 43PECh. R3 - Prob. 44PECh. R3 - Prob. 45PECh. R3 - Prob. 46PECh. R3 - Prob. 47PECh. R3 - Prob. 48PECh. R3 - Prob. 49PECh. R3 - Prob. 50PECh. R3 - Prob. 51PECh. R3 - Prob. 52PECh. R3 - Prob. 53PECh. R3 - Prob. 54PECh. R3 - Prob. 55PECh. R3 - Prob. 56PECh. R3 - Prob. 57PECh. R3 - Prob. 58PECh. R3 - Prob. 59PECh. R3 - Prob. 60PECh. R3 - Prob. 61PECh. R3 - Prob. 62PECh. R3 - Prob. 63PECh. R3 - Prob. 64PECh. R3 - Prob. 65PECh. R3 - Prob. 66PECh. R3 - Prob. 67PECh. R3 - Prob. 68PECh. R3 - Prob. 69PECh. R3 - Prob. 70PECh. R3 - Prob. 71PECh. R3 - Prob. 72PECh. R3 - Prob. 73PECh. R3 - Prob. 74PECh. R3 - Prob. 75PECh. R3 - Prob. 76PECh. R3 - Prob. 77PECh. R3 - Prob. 78PECh. R3 - Prob. 79PECh. R3 - Prob. 80PECh. R3 - Prob. 81PECh. R3 - Prob. 82PECh. R3 - Prob. 83PECh. R3 - Prob. 84PECh. R3 - Prob. 85PECh. R3 - Prob. 86PECh. R3 - Prob. 87PECh. R3 - Prob. 88PECh. R3 - For Exercises 69-96, simplify by using the order...Ch. R3 - Prob. 90PECh. R3 - Prob. 91PECh. R3 - Prob. 92PECh. R3 - Prob. 93PECh. R3 - Prob. 94PECh. R3 - Prob. 95PECh. R3 - Prob. 96PECh. R3 - Prob. 97PECh. R3 - Prob. 98PECh. R3 - Prob. 99PECh. R3 - Prob. 100PECh. R3 - Prob. 101PECh. R3 - Prob. 102PECh. R3 - Prob. 103PECh. R3 - Prob. 104PECh. R3 - Prob. 105PECh. R3 - Prob. 106PECh. R3 - Prob. 107PECh. R3 - Prob. 108PECh. R3 - Prob. 109PECh. R3 - Prob. 110PECh. R3 - Prob. 111PECh. R3 - Prob. 112PECh. R3 - Prob. 113PECh. R3 - Prob. 114PECh. R3 - Prob. 115PECh. R3 - Prob. 116PE
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