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Sunny_sXe [5.5K]
2 years ago
9

18. A science teacher has a supply of 20% saline solution and a supply of 50% saline solution. How much of each solution should

the teacher mix together to get 60 mL of 28% saline solution for an experiment? (1 point)
A) 16 mL of the 20% solution and 44 mL of the 50% solution

B) 44 mL of the 20% solution and 16 mL of the 50% solution

C) 16 mL of the 20% solution and 16 mL of the 50% solution

D) 44 mL of the 20% solution and 44 mL of the 50% solution
Mathematics
1 answer:
VladimirAG [237]2 years ago
3 0
The answer is (c)16 mL of the 20% solution and the 16 mLof the 50% solution
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8.5b−3.4∙(13a−3.2b)+a
Nesterboy [21]

Answer:

8.5b - 3.4(13a - 3.2b) + a = 19.4b - 43.2a

Step-by-step explanation:

It is a simple mathematical problem with multiple like terms. We can solve it by applying basic mathematical rules of multiplication and addition/subtration.

8.5b - 3.4(13a - 3.2b) + a

= 8.5b - 3.4*13a -3.4*(-3.2b) + a

= 8.5b - 3.4*13a + 3.4*3.2b + a

= 8.5b - 44.2a + 10.88 b + a

Now, only like terms can be added to each other

= (8.5b + 10.9b) + (a - 44.2a)

= 19.4b + (-43.2a)

= 19.4b - 43.2a    

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