Answered

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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.
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