Intermediate Algebra Tutorial 4


Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 4: Operations on Numbers


WTAMU > Virtual Math Lab > Intermediate Algebra > Tutorial 4: Operations on Numbers


 

checkAnswer/Discussion to 1a

 -5 + (-16)
 

-5 + (-16) = -21. 

Note that we are adding two numbers with the same sign.  The sum of the absolute values 5 and 16 is 21 and their common sign is negative. Therefore, our answer is -21.

(return to problem 1a)


 

checkAnswer/Discussion to 1b

7 + (-3)
 

7 + (-3) = 4. 

Note that we are adding two numbers that have opposite signs.  The difference between the absolute values of 7 and 3 is 4 and the sign of the number with the larger absolute value (7) is positive.  Therefore, our answer is 4.

(return to problem 1b)


 

checkAnswer/Discussion to 1c

 -10 - (-2)

-10 - (-2) = -10 + 2 = -8. 

Subtracting a negative 2 is the same as adding a positive 2.  Once we change it to addition, we now are adding two numbers with opposite signs.  The difference of the absolute values of 10 and 2 is 8 and the sign of the larger absolute value (-10) is negative.  Therefore, our answer is -8.

(return to problem 1c)


 

checkAnswer/Discussion to 2a

 (-4)(9)
 

 (-4)(9) = -36. 

We are multiplying two numbers with opposite signs.  The product of 4 and 9 is 36 and multiplying with opposite signs gives us a negative. Therefore, our answer is -36.

(return to problem 2a)


 

checkAnswer/Discussion to 2b

problem 2b
 

problem 2b= .5 or ½. 

We are dividing two numbers with the same sign.  The quotient of 3 and 6 is .5 and dividing with the same signs gives us a positive number. Therefore, our answer is .5.

(return to problem 2b)


 

checkAnswer/Discussion to 3a

problem 3a
 

ad3a

*Write the base 3 in a product 4 times
*Multiply 

 
(return to problem 3a)


 

checkAnswer/Discussion to 3b

problem 3b
 

ad3b

*Write the base -10 in a product 2 times
*Multiply

 
(return to problem 3b)


 

checkAnswer/Discussion to 3c

problem 3c
 

ad3c
*Negate 10 squared
*Put a - in front of 10 written in a product 2 times
*Multiply

 
(return to problem 3c)


 

checkAnswer/Discussion to 4a

problem 4a
 

ad4a

The thought behind this is that we are looking for the square root of 36, which means we are looking for a number that when we square it we get 36.   Since 6 squared is 36, 6 is the square root of 36.  Note that we are only interested in the principal root and since 36 is positive and there is not a sign in front of the radical, our answer is positive 6.  If there had been a negative in front of the radical our answer would have been -6

 (return to problem 4a)


 

checkAnswer/Discussion to 4b

problem 4b
 

ad4b

Now we are looking for the fourth root of 1/16, which means we are looking for a number that when we take it to the 4th power, we get 1/16.  Since 1/2 to the 4th power is 1/16, our answer is going to be 1/2.

(return to problem 4b)


 

checkAnswer/Discussion to 5a

problem 5a
 

ad5a

*(  ) 
*Exponents
*Multiplication
*Subtraction

 
(return to problem 5a)


 

checkAnswer/Discussion to 5b

problem 5b
 

ad5b

 

*Absolute value 
*(fancy grouping symbol)
*Square root 
 

*Subtraction in numerator

*Divide
 
 

(return to problem 5b)


 

checkAnswer/Discussion to 6a

Find the value of the expression when a = 3 and b = -2.
problem 6a
 

ad6a

*Replace a with 3 and b with -2
*Multiply within (  )
*Subtract within (  )
*Multiply

 
(return to problem 6a)

 

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WTAMU > Virtual Math Lab >Intermediate Algebra > Tutorial 4: Operations on Numbers


Last revised on June 10, 2011 by Kim Seward.
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