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3 years ago

Factor completely 2x^2 − 2x − 40. 2(x − 5)(x 4) (2x − 10)(x 4) (x − 5)(2x 8) 2(x − 4)(x 5)

Mathematics

1 answer:

Zina [86]3 years ago

3 0

2x^2 - 2x - 40 = 2x^2 -10x + 8x - 40 = 2x(x - 5) + 8(x - 5)

= (x-5)(2x+8) = 2(x-5)(x+4)

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2 years ago

❊ Simplify :

Nataliya [291]

To simplify the given expression .

$\red{\frak{Given}}\Bigg\{ \sf \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 }$

$\rule{200}4$

$\sf\longrightarrow \small \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 } \\\\\\\sf\longrightarrow \small \dfrac{ x-1}{x^2-x -2x +2} +\dfrac{ x-2}{x^2-3x-2x+6} +\dfrac{ x -5}{x^2-5x -3x + 15 } \\\\\\\sf\longrightarrow\small \dfrac{ x -1}{ x ( x - 1) -2(x-1) } +\dfrac{ x-2}{x ( x -3) -2( x -3)} +\dfrac{ x -5}{ x(x-5) -3( x -5) } \\\\\\\sf\longrightarrow \small \dfrac{ x -1}{ ( x-2) (x-1) } +\dfrac{ x-2}{( x -2)(x-3) } +\dfrac{ x -5}{ (x-3)(x-5) } \\\\\\\sf\longrightarrow\small \dfrac{ 1}{ x-2} +\dfrac{ 1}{ x -3} +\dfrac{1}{ x -3} \\\\\\\sf\longrightarrow \small \dfrac{1}{x-2} +\dfrac{2}{x-3} \\\\\\\sf\longrightarrow \small \dfrac{ x-3 +2(x-2)}{ ( x -3)(x-2) } \\\\\\\sf\longrightarrow \small \dfrac{ x - 3 +2x -4 }{ (x-3)(x-2) } \\\\\\\sf\longrightarrow \underset{\blue{\sf Required \ Answer }}{\underbrace{\boxed{\pink{\frak{ \dfrac{ 3x -7}{ ( x -2)(x-3) } }}}}}$

$\rule{200}4$

5 0

2 years ago

Read 2 more answers

Please help me with this!!!

Dafna1 [17]

Answer:

option C is correct answer ..

Step-by-step explanation:

angle A + angle B + angle C = 180° ( by angle sum property of triangle )

3x + 4x-19+ 3x -1 = 180

10x + -20 = 180

10x = 180 +20 = 200

x = 200/10 = 20 °

angle A = 3× 20 = 60 °

angle B = 4× 20 - 1 9 = 61 °

angle C = 3× 20 -1 = 59 °

angle B is greatest so side opposite to it will be greatest in length ....so length of AC is greatest ....

so option C is the correct answer of this question ...

plz mark my answer as brainlist plzzzz vote me also as I have done this question by taking a lot of time Hope it will be helpful for you ....

5 0

2 years ago