Name two solutions for each inequality. A. n \geq 3 11/16+ 4 1/2 B. k< 6 2/5 * 15

Question

Name two solutions for each inequality.

A. n
`\geq` 3 11/16+ 4 1/2


















































B. k< 6 2/5 * 15

Answer 1

So,

#1:
`n \geq 3 (11)/(16) + 4 (1)/(2)`

Convert to like improper fractions.

`n \geq (59)/(16) + (72)/(16)`

Add.

`n \geq (131)/(16)\ or\ 8 (3)/(16)`

So, one solution could be
`8 (3)/(16)`.

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


#2:
`k \ \textless \ 6 (2)/(5) * 15`

Convert into improper fraction form.

`k \ \textless \ (32)/(5) * 15`

Multiply.

`((2^5)(3)(5))/(5)`

Cross-cancel, and we have our final result.

`(2^5)(3) = 96`
k < 96

96 is not a solution.

95 is a solution.

So is 94, 93, 92, etc, and all numbers in between.