Answered

Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Name two solutions for each inequality.



 







A. n [tex] \geq [/tex] 3 11/16+ 4 1/2


















































B. k< 6 2/5 * 15

Sagot :

SchwarzschildRadius
So,

#1: [tex]n \geq 3 \frac{11}{16} + 4 \frac{1}{2} [/tex]

Convert to like improper fractions.
[tex]n \geq \frac{59}{16} + \frac{72}{16} [/tex]

Add.
[tex]n \geq \frac{131}{16}\ or\ 8 \frac{3}{16} [/tex]

So, one solution could be [tex]8 \frac{3}{16} [/tex].

Another solution could by 9.  There is also 10, 11, 12, etc., and all numbers in between.


#2: [tex]k \ \textless \ 6 \frac{2}{5} * 15[/tex]

Convert into improper fraction form.
[tex]k \ \textless \ \frac{32}{5} * 15[/tex]

Multiply.
[tex] \frac{(2^5)(3)(5)}{5} [/tex]

Cross-cancel, and we have our final result.
[tex](2^5)(3) = 96[/tex]
k < 96

96 is not a solution.

95 is a solution.

So is 94, 93, 92, etc, and all numbers in between.
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.