19. A newborn baby chicken weighs 3 pounds. If an adult chicken can weigh up to 43 times a newborn chicken. then it would result to 43 pounds. Your teacher asked you on how did you arrived at your answer. What is the best explanation you can give? A. First. make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression to and finally multiplying the fraction by 43 B. First. make the exponent positive by moving the negative exponent to the numerator resulting to then simplify the expression to and finally multiplying the fraction by 43 C. First. make the exponent positive by moving the negative exponent to the denoninator resulting to then simplify the expression to and finally multiplying the fraction by 43 D. First. make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression to and finally dividing the fraction by 43

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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19. A newborn baby chicken weighs 3 pounds. If an adult chicken can weigh up to 43 times a newborn chicken, then it would result to
pounds. Your teacher asked you on how did you arrived at your answer. What is the best explanation you can give?
A. First, make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression
34
to - and finally multiplying the fraetion by 43
B. First. make the exponent positive by moving the negative exponent to the numerator resulting to then sinmplify the expression to
and finally multiplying the fraction by 43
C. First. make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression
and finally multiplying the fraction by 43
to
D. First. make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression
to - and finally dividing the fraction by 43
20. If a 36 - storey building is 110 meter high, using the formula /3h/2 for sight distance where d is the distance in miles and h is the height
where the person is, how far can you see the building on a clear day? (1 mile = 1609.3 m). How would you illustrate the problem? If the light
of the building inerease/decreases. what might happen to the sight distance?
A. If the height of the building inereases/decreases, the sight distance might go farther nearer respectively.
В.
If the height of the building increases/decreases, the sight distance might go nearer farther respectively.
C. If the height of the building increases/decreases, the sight distance might go bevond the building.
Transcribed Image Text:19. A newborn baby chicken weighs 3 pounds. If an adult chicken can weigh up to 43 times a newborn chicken, then it would result to pounds. Your teacher asked you on how did you arrived at your answer. What is the best explanation you can give? A. First, make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression 34 to - and finally multiplying the fraetion by 43 B. First. make the exponent positive by moving the negative exponent to the numerator resulting to then sinmplify the expression to and finally multiplying the fraction by 43 C. First. make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression and finally multiplying the fraction by 43 to D. First. make the exponent positive by moving the negative exponent to the denominator resulting to then simplify the expression to - and finally dividing the fraction by 43 20. If a 36 - storey building is 110 meter high, using the formula /3h/2 for sight distance where d is the distance in miles and h is the height where the person is, how far can you see the building on a clear day? (1 mile = 1609.3 m). How would you illustrate the problem? If the light of the building inerease/decreases. what might happen to the sight distance? A. If the height of the building inereases/decreases, the sight distance might go farther nearer respectively. В. If the height of the building increases/decreases, the sight distance might go nearer farther respectively. C. If the height of the building increases/decreases, the sight distance might go bevond the building.
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