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sesenic [268]
3 years ago
14

Can i get help wit these? i need 5 7 and 8

Mathematics
1 answer:
telo118 [61]3 years ago
6 0

Answer:

<5 = 132°

<7 = 132°

<8 = 48°

Step-by-step explanation:

aAgle 2 and angle 5 are supplementary which means their sum is 180° so angle 5 is 180 - 48 = 32°

Angle 5 and angle 7 are congruent so their measures are equal 132°

Angle 8 and angle 2 are congruent and equal to each other as well

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Answer:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

Step-by-step explanation:

Required

Simplify

Solving (1):

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}

Factorize the numerator and the denominator

\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}

Factor out x+2 at the numerator

\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}

Express x^2 - 9 as difference of two squares

\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}

\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}

Expand the denominator

\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}

Factorize

\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}

\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}

Cancel out same factors

\frac{(x+3)}{x}

Hence:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

Solving (2):

\frac{3x^2 - 5x - 2}{x^3 - 2x^2}

Expand the numerator and factorize the denominator

\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}

Factorize the numerator

\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}

Factor out x - 2

\frac{(3x + 1)(x - 2)}{x^2(x- 2)}

Cancel out x - 2

\frac{3x + 1}{x^2}

Hence:

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

Solving (3):

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}

Express x^2 - 9 as difference of two squares

\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}

Factorize all:

\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}

Cancel out x + 3 and 3 + x

\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}

\frac{3 - x}{x- 3} * \frac{5}{2x}

Express 3 - x as -(x - 3)

\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\

-1 * \frac{5}{2x}

-\frac{5}{2x}

Hence:

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

Solving (4):

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}

Expand x^2 - 6x + 9 and factorize 5x - 15

\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}

Factorize

\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}

\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}

Cancel out x - 3

\frac{(x -3)}{5} / \frac{5}{3-x}

Change / to *

\frac{(x -3)}{5} * \frac{3-x}{5}

Express 3 - x as -(x - 3)

\frac{(x -3)}{5} * \frac{-(x-3)}{5}

\frac{-(x-3)(x -3)}{5*5}

\frac{-(x-3)^2}{25}

Hence:

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

Solving (5):

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}

Factorize the numerator and expand the denominator

\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}

Factor out x - 1 at the numerator and factorize the denominator

\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}

x +1

Hence:

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

Solving (6):

\frac{9x^2 + 3x}{6x^2}

Factorize:

\frac{3x(3x + 1)}{3x(2x)}

Divide by 3x

\frac{3x + 1}{2x}

Hence:

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

Solving (7):

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}

Change / to *

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Expand

\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Factorize

\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

Cancel out x - 2 and x - 1

\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}

Cancel out x

\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}

\frac{12x^2}{4x}

3x

Hence:

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

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