3 Title

Beginning Algebra
Answer/Discussion to Practice Problems
Tutorial 7: Multiplying and Dividing Real Numbers

 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.

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

 (0)(-100) = 0.

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

 Working it left to right we get:

 (-2)(3)(5) =  (-6)(5) = -30

 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.

 Answer/Discussion to 2b

 7/0 = undefined.

 Answer/Discussion to 2c

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

 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.

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

 Answer/Discussion to 4a

 Plugging 5 for x and -5 for y and simplifying we get:

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

Last revised on July 25, 2011 by Kim Seward.
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