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4 years ago

how much water should be mixed with 6 gallons of 55% solution of sulfuric acid to make a 20% solution of sulfuric acid

1 answer:

6 0

Hello.

For all dilution problems use the dilution equation:
C1V1 = C2V2 
20*V1 = 6*55 
V1 = 6*55/20 
V1 = 16.5 gallons 
The final volume must be 16.5 gallons 
You must add 16.5 - 6 = 10.5 gallons water to be added.

Have a nice day

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Q1: Identify the graph of the equation and write an equation of the translated or rotated graph in general form.

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Answer:

C. hyperbola; $9x^2-25y^2-250y-850=0$

Step-by-step explanation:

The given conic has equation:

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Divide through by 225.

$\frac{9x^2}{225}-\frac{25y^2}{225}=\frac{225}{225}$

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The hyperbola has been translated from the origin to (0,5).

The translated hyperbola will have equation;

$\frac{(x-0)^2}{25}-\frac{(y-5)^2}{9}=1$

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Expand

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Rewrite in general form;

$9x^2-25y^2+250y-625-225=0$

$9x^2-25y^2+250y-850=0$

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