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3 years ago

1. A person spends1/4 -part of his

Mathematics

2 answers:

Ksivusya [100]3 years ago

6 0

Answer:

*2520

Step-by-step explanation:

A quarter of him income was spent on food

X (representing his income) - x/4= 3x/4

Then 2/3 was spent on house rent. Notice the, the question didnt say 2/3 of his income. Therefore it would be 2/3 of what is remaining after he pays for food.

2/3 * 3x/4= x/2

3x/4-x/2=x/4.

The remaining after all expenses are paid is 630. So:

x/4=630

x=630*4

x=2520

Valentin [98]3 years ago

5 0

1/4 + 2/3 + 1/12 = 1

That is fraction on

Food + house rent + other items = Total amount

If 1/12 = 630.............. equation 1

Then 1/4 + 2/3 = 11/12

11/12= x................. equation 2

Cross multiply equation 1 & 2

1/12*x = 1155/2

To find x, cross multiply again

2x = 13860

x = 13860/2

x = 6930

This means that 11/12 + 1/12 = 6930 + 630

= 7560

Therefore 2/3(fraction for house rent)*7560(total amount)=5040

Answer is 5040

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3 years ago

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

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3 years ago