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Komok [63]
3 years ago
5

A tank has a capacity of 10 gallons. When it is full it contains 15% alcohol. How many gallons must be replaced by an 80% soluti

on to goive 10 gallons of a 70% solution?
Mathematics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer:

8.46 gallons needed to be replaced

Step-by-step explanation:

Let x be the alcohol amount that replaced

-Original has concentration of 15% multiply by the amount of 10 ,which is equal to (0.15×10).

-Removed has concentration of 15% multiply by the amount of x ,which is equal to (0.15 × X).

-Added has concentration of 80% multiply by the amount of X ,which is equal to (0.8× X).

-Solution has concentration of 70% multiply by the amount of 10 ,which is equal to (0.70×10).

So that, the equation will be:

Solution=Original + added - removed

7=1.5+0.8x-0.15x

7-1.5=0.65x

0.65x=55

x=55/0.65

x=8.46

8.46 gallons needed to be replaced.

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-9-8(1+4h) = -17. We need to solve for h.
First, use the distributive property for 8(1+4h):
8(1+4h) = 8*1 + 8*4h = 8 + 32h

So -9-8(1+4h) = -17
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Add 17 on both sides to have the variables on a side and the numbers on the other:
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divide both sides by -32 to get the variable h alone and its value on the other side:
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So -9-8(1+4h) = -17 for h = 0.

You can recheck your answer (very important):
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The answer has been approved.

Hope this Helps! :)
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