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3 years ago

What are the zeros of the function y = x2 + 10x – 171, and why?

Mathematics

2 answers:

katen-ka-za [31]3 years ago

8 0

Hello,

Answer B

y=x²+10x-171=x²-9x+19x-171=x(x-9)+19(x-9)=(x-9)(x+19)

aleksandr82 [10.1K]3 years ago

4 0

Y = x^2 + 10x - 171
y = (x - 9)(x + 19)

x - 9= 0 x + 19 = 0
x = 9 x = -19

Answer B covers all requirements... the factored form is
<span>y= (x + 19)(x - 9) </span>
<span>and the zeros are -19 and 9</span>

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2 years ago

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3 years ago

Read 2 more answers

I need help on all of these

lord [1]

well #7 is a solution.

y= 9x (4,36)

9 times 4 = 36

36/9=4

so the ordered pair is a solution.

y=3x-5 (7,16)

3 times 7 -5= 16

so once again the ordered pair is a solution.

Hoped this helped.

5 0

3 years ago

❊ Simplify :

Nataliya [291]

To simplify the given expression .

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

$\rule{200}4$

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