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Alenkinab [10]
2 years ago
6

What would the concentration be if we mixed 200g of 10% sugar solution and 300g of 20% sugar solution?

Mathematics
2 answers:
Irina18 [472]2 years ago
7 0
The new concentration is obtained by
\frac{200 \times 10 + 300 \times 20}{200 + 300} = \frac{2000 + 6000}{500} = \frac{7000}{500} = 14\%
stepan [7]2 years ago
3 0
200 g + 300 g = 500 g
200 * 0.1 + 300 * 0.2 = 500 * x
20 + 60 = 500 * x
80 = 500 * x
x = 80 : 500
x = 0.16
Answer: The concentration would be 16 % ( it must be closer to 20% than to 10% ).
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<u>Given:</u>

The given equation is \log _{4}(x)+\log _{4}(x-3)=\log _{4}(-7 x+21)

We need to determine the extraneous solution of the equation.

<u>Solving the equation:</u>

To determine the extraneous solution, we shall first solve the given equation.

Applying the log rule \log _{c}(a)+\log _{c}(b)=\log _{c}(a b), we get;

\log _{4}(x(x-3))=\log _{4}(-7 x+21)

Again applying the log rule, if \log _{b}(f(x))=\log _{b}(g(x)) then f(x)=g(x)

Thus, we have;

x(x-3)=-7 x+21

Simplifying the equation, we get;

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Factoring the equation, we get;

(x-3)(x+7)=0

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The extraneous solutions are the solutions that does not work in the original equation.

Now, to determine the extraneous solution, let us substitute x = 3 and x = -7 in the original equation.

Thus, we get;

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Since, we know that \log _{a}(0) is undefined.

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Undefined = Undefined

This is false.

Thus, the solution x = 3 does not work in the original equation.

Hence, x = 3 is an extraneous solution.

Similarly, substituting x = -7, in the original equation. Thus, we get;

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Hence, x = -7 is an extraneous solution.

Therefore, the extraneous solutions are x = 3 and x = -7

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Before the second break, it was painted:
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Now, we will sum what he painted for now:
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When the painter takes his fifth break, there will be <span>141.35 ft² of the wall painted.</span>
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