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Answer/Discussion to 1a (-2)(-25)

 

(-2)(-25) = 50. The product of the absolute values 2 and 25 is 50 and they have the same sign, so that is how we get the answer 50.

 

(return to problem 1a)

 

 

Answer/Discussion to 1b (0)(-100)

 

(0)(-100) = 0.

 

(return to problem 1b)

 

 

Answer/Discussion to 1c (-2)(3)(5)

 

Working it left to right we get:

 

(-2)(3)(5) =  (-6)(5) = -30

 

(return to problem 1c)

 

 

Answer/Discussion to 2a

 

(-25)/(5) = -5.  The quotient of the absolute values (25)/(5) = 5 and they have opposite signs, so that is how we get the answer -5.

 

(return to problem 2a)

 

 

Answer/Discussion to 2b

 

7/0 = undefined.

 

(return to problem 2b)

 

 

Answer/Discussion to 2c

 

*Div. is the same as mult. by reciprocal *Mult. num. together *Mult. den. together *(-)(-) = - *Reduce fraction

 

(return to problem 2c)

 

 

Answer/Discussion to 3a

 

Since we have several operations going on in this problem, we will have to use the order of operations to make sure that we get the correct answer.  If you need to review the order of operations go to Tutorial 4: Operations of Real Numbers.

 

*Evaluate inside the absolute values     *Multiply   *Add *Reduce fraction

 

(return to problem 3a)

 

 
 

Answer/Discussion to 4a

 

To review evaluating an expression go to Tutorial 4: Introduction to Variable Expressions and Equations. Plugging 5 for x and -5 for y and simplifying we get:

 

*Plug in 5 for x and -5 for y *Exponent *Multiply *Subtract

 

(return to problem 4a)

 

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Last revised on July 25, 2011 by Kim Seward.
All contents copyright (C) 2001 - 2011, WTAMU and Kim Seward. All rights reserved.