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jolli1 [7]
2 years ago
7

Please help me with this!!! ​

Mathematics
1 answer:
Strike441 [17]2 years ago
7 0

Answer:

See explanation below

Step-by-step explanation:

Let's use Q10 as an example and you will work on Q11-12 as they are the same.

I believe the middle column (5x-4) wants you to substitute the given x, so:

Q10.1: 5(-2)-4

Q10.2: 5(4)-4

Q10.3: 5(0)-4

Q10.4: 5(1)-4

The right most column (f(x)) should be asking you the answers from your middle column, so:

Q10.1: -14

Q10.2: 16

Q10.3: -4

Q10.4: 1

As for the domain and range, let's define these two names in case you haven't learnt them:

Domain: What reasonable x values can be inputted into the function.

Range: What reasonable results will be shown after calculating.

In these case, you can input any real numbers into x as you like into the function, the results will always be other real numbers.

Hence, for Q10-12, D and R should be "All real numbers"

Just some extra info, Let the function f(x) be \frac{1}{x+1}. Since the denominator cannot be 0 in a fraction, we say that the domain of x is "All real numbers except -1" as -1+1 = 0 and \frac{1}{0} is undefined.

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svlad2 [7]

Answer:

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Step-by-step explanation:

The question is unreadable, however the real polynomial is:

The polynomial fraction is:

P = \frac{9x + 11}{6x^2 + 23x + 21}

And the decomposition is:

\frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}

The solution is as follows:

P = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}

Substitute the expression for P

\frac{9x + 11}{6x^2 + 23x + 21} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}

Expand the numerator of the polynomial

\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}

Take LCM

\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)*(3x + 7) + g(x)*(2x + 3)}{(2x + 3)(x + 7)}

Cancel out both denominators

9x + 11} = f(x)*(3x + 7) + g(x)*(2x + 3)

Represent f(x) as A and g(x) as B

9x + 11} = A*(3x + 7) + B*(2x + 3)

Open bracket

9x + 11} = 3Ax + 7A + 2Bx + 3B

9x + 11} = 3Ax + 2Bx+ 7A  + 3B

9x + 11} = (3A + 2B)x+ 7A  + 3B

By comparison:

3A + 2B = 9 ---- (1)

7A  + 3B =11 ---- (2)

Make B the subject in (1)

B = \frac{9 - 3A}{2}

Substitute B = \frac{9 - 3A}{2} in (2)

7A + 3(\frac{9 - 3A}{2}) = 11

Multiply through by 2

2*7A +2* 3(\frac{9 - 3A}{2}) = 11*2

14A + 3(9 - 3A) = 22

14A + 27 - 9A = 22

Collect Like Terms

14A - 9A = 22-27

5A = -5

A = \frac{-5}{5}

A = -1

Recall that:

B = \frac{9 - 3A}{2}

B = \frac{9 - 3 * -1}{2}

B = \frac{9 + 3}{2}

B = \frac{12}{2}

B = 6

A = -1 and B = 6

3A + 2B = 9 ---- (1)

7A  + 3B =11 ---- (2)

So:

f(x) = A

f(x) = -1

g(x) =B

g(x) =6

And the decomposition is:

\frac{-1}{2x + 3} + \frac{6}{3x + 7}

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