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Anna11 [10]
3 years ago
10

Chase has 3 gallons of a solution that is 30% antifeeze that he wants to use to winterize his car. How much pure antifreeze shou

ld he add to this solution to produce a solution that is 65% antifreeze.
Mathematics
1 answer:
makvit [3.9K]3 years ago
8 0

Let  x gallons be the amount of pure antifreeze that should be added to the 30% solution to produce a solution that is 65% antifreeze. Then the total amount of antifreeze solution will be x+3 gallons.

There are 30% of pure antifreeze in 3 gallons of solution, then

3 gallons - 100%,

a gallons - 30%,

where a gallons is the amount of pure antifreeze in given solution.

Mathematically,

\dfrac{3}{a}=\dfrac{100}{30},\\ \\a=\dfrac{3\cdot 30}{100}=0.9\ gallons.

Now in new solution there will be x+0.9 gallons of pure antefreeze.

x+3 gallons - 100%,

x+0.9 - 65%

or

\dfrac{x+3}{x+0.9}=\dfrac{100}{65},\\ \\65(x+3)=100(x+0.9),\\ \\65x+195=100x+90,\\ \\35x=105,\\ \\x=3\ gallons.

Answer: he should add 3 gallons of pure antifreeze.

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Answer:

2/3

Step-by-step explanation:

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Does that help?

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mylen [45]

Answer:

56.39 nm

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where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm

Hence the thickness difference should be

t_1-t_2=131.58-75.19=56.39 \ nm

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the answer is C. because it is the only one that does not contain a common factor :)


3 0
3 years ago
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