When you identify what goals are conflicting, it is a part of

Mathematics

1 answer:

olga_2 [115]2 years ago

3 0

Answer: Analysis

Step-by-step explanation:

When analysing a project, one must try to find out everything that could impart on the success of the project such that any problems can be fixed and any potential problems be made contingencies for.

This is why it is here that you check to see whether and which goals are conflicting so that they can be fixed or aligned to ensure that the project comes out as a success.

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Resolve into partial fractions 7-5x/2x^2+x-1

Flauer [41]

The decomposition of partial fractions is to start with the simplified reply and then take it apart, to "decompose" the final expression into its initial polynomial fractions.

Given:

$\to \bold{\frac{7-5x}{2x^2+x-1}}\\\\$

To find:

partial fractions=?

Solution:

$\to \bold{\frac{7-5x}{2x^2+x-1}}\\\\\to \bold{\frac{7-5x}{2x^2+x(2-1)-1}}\\\\\to \bold{\frac{7-5x}{2x^2+2x-x-1}}\\\\\to \bold{\frac{7-5x}{2x(x+1)-1(x+1)}}\\\\\to \bold{\frac{7-5x}{(2x-1)(x+1)} = \frac{A}{(2x-1)} - \frac{B}{(x+1)} }\\\\\to \bold{7-5x = \frac{A ((2x-1)(x+1))}{(2x-1)} - \frac{B((2x-1)(x+1))}{(x+1)} }\\\\\to \bold{7-5x = A(x+1) -B(2x-1) }\\\\$

putting $x=-1$

$\to \bold{7-5(-1) = A(-1+1) -B(2(-1)-1) }\\\\\to \bold{7+5 = A(0) -B(-2-1) }\\\\\to \bold{12 = +3B }\\\\\to \bold{B = \frac{12}{3} }\\\\\to \bold{B = 4 }\\\\$

putting $x= \frac{1}{2}$

$\to \bold{7-5(\frac{1}{2}) = A(\frac{1}{2}+1) -B(2(\frac{1}{2})-1) }\\\\\to \bold{7-\frac{5}{2} = A(\frac{3}{2}) -B((\frac{2}{2})-1) }\\\\\to \bold{7-\frac{5}{2} = A(\frac{3}{2}) -B(1-1) }\\\\\to \bold{\frac{14-5}{2} = A(\frac{3}{2}) -B(0) }\\\\\to \bold{\frac{9}{2} = A(\frac{3}{2})}\\\\\to \bold{\frac{9}{2} \times \frac{2}{3} = A}\\\\\to \bold{\frac{9}{3} = A}\\\\\to \bold{A=3}\\\\$