College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Are you Ready for College Algebra? - A. Diagnostic Test:: Real Numbers and Exponents: 7


(a) 8$\mathcal{x}$$^{3}$$\mathcal{y}$$^{5}$ (b) $\frac{25}{a^3}$ (c) $\mathcal{y}$

Work Step by Step

a) all numbers inside parentheses are already simplified so brackets can be removed and all 'like terms' multiplied. When multiplying like terms, the exponents are added together. Where, x^{2} multiplied by x^{1} becomes x^{3} and y^{3} multiplied by y^{2} becomes y^{5}. The constants (4 and 2) are simply multiplied using 'regular' multiplication to get 8. b) Since 2 is outside the parentheses, the exponents of all the terms inside them must be multiplied by 2. Where, for the numerator; 5^{1}, 1 x 2 forms 5^{2}, and for a^{\frac{1}{2}, where \frac{1}{2} x 2 becomes a^{1} and for the denominator of a^{2}, 2 x 2 becomes a^{4}. Since there are now 'unsimplified' terms in the numerator and denominator working is incomplete, So the 5^{2} becomes 25 in the numerator and to express the answer in positive index form, the numerator's 'a' exponent is subtracted from the denominator (4-1) to form a^{3} c) Similiar to part b the parentheses of the second collection of terms needs to be simplified, by multiplying the exponents of everything inside the brackets by 2 to form (x^{2}y^{4}). Since the first collection of terms has negative exponents, they need to 'inverted' to form a fraction. Doing this will result in a fraction, with 1 x (x^{2}y^{4}) forming the numerator and x^{2} and y^{3} in the denominator. Using the Index rules, the exponents will be subtracted from one another, and to form positive indices x^{2} \div x^{2} form an exponent of 0 which equates to 1. And y^{3} \div y^{2} form y^{1}, which is the final answer.
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