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wlad13 [49]
4 years ago
12

Interest

Mathematics
1 answer:
aleksandr82 [10.1K]4 years ago
3 0

i = 2400 \times .010 \div 20
i = 24 \times 20 = 480
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The following trinomial is completely factored: 12x2 - 15x + 24
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For each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to
kiruha [24]

Answer:

D) 11/3

Step-by-step explanation:

Let x represent monthly salary.

We have been given that for each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity.        

Salary earned in 11 months would be 11x.

Money donated in 11 months would be \frac{11x}{10}.

Further we are told that in December, the worker earned N times his usual monthly salary and donated one-fifth of his earnings to charity.

Salary for December would be Nx.

Money donated in December would be \frac{Nx}{5}.  

The worker's charitable contributions totaled one-eighth of his earnings for the entire year that is \frac{1}{8}\cdot (11x+Nx).

\frac{11x}{10}+\frac{Nx}{5}=\frac{1}{8}\cdot (11x+Nx)

Dividing whole equation by x, we will get:

\frac{11x}{10x}+\frac{Nx}{5x}=\frac{1}{8x}\cdot x(11+N)                

\frac{11}{10}+\frac{N}{5}=\frac{1}{8}\cdot (11+N)

\frac{11}{10}+\frac{N}{5}=\frac{11}{8}+\frac{N}{8}

Combine like terms:

\frac{N}{5}-\frac{N}{8}=\frac{11}{8}-\frac{11}{10}

\frac{N}{5}*40-\frac{N}{8}*40=\frac{11}{8}*40-\frac{11}{10}*40

8N-5N=11*5-11*4

3N=55-44

3N=11

\frac{3N}{3}=\frac{11}{3}\\\\N=\frac{11}{3}

Therefore, the value of N is 11/3 and option D is the correct choice.

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Answer:

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Step-by-step explanation:

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multiply (5x^2 x − 4)(x 2). 5x^3 x^2 − 4x − 8 5x^3 11x^2 − 2x − 8 5x^3 10x^2 − 4x − 8 5x^3 20x^2 2x − 8
Ulleksa [173]
(5x^2 + x - 4)(x + 2) = x(5x^2 + x - 4) + 2(5x^2 + x - 4) = 5x^3 + x^2 - 4x + 10x^2 + 2x - 8 = 5x^3 + 11x^2 - 2x - 8
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