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

 (-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.

 7/0 = undefined.

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

 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

 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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