Please help me on this now.

Mathematics

1 answer:

alexandr1967 [171]2 years ago

3 0

Answer:

the frist one is 2 the second on is 3.5 and the third one is 3 and the last on is 2.9

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Name two solutions for each inequality.

ahrayia [7]

So,

#1: $n \geq 3 \frac{11}{16} + 4 \frac{1}{2}$

Convert to like improper fractions.
$n \geq \frac{59}{16} + \frac{72}{16}$

Add.
$n \geq \frac{131}{16}\ or\ 8 \frac{3}{16}$

So, one solution could be $8 \frac{3}{16}$ .

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

#2: $k \ \textless \ 6 \frac{2}{5} * 15$

Convert into improper fraction form.
$k \ \textless \ \frac{32}{5} * 15$

Multiply.
$\frac{(2^5)(3)(5)}{5}$

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.

6 0

3 years ago

Shay mal e 8 2/3 cups of snacks mix she puts all the snack mix into plastic bags she puts 1/3 cup of the snack mix in each bag h

AnnyKZ [126]

Answer:

She needs 26 plastic bags.

Step-by-step explanation:

Given:

Amount of cups of snacks mix = 8 2/3

Rewriting 8 2/3 we get,

Amount of snacks mix = $\frac{26}{3}$