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Virty [35]
3 years ago
10

Find the constant of proportionality for the table and write in the form y = kx.

Mathematics
2 answers:
Elden [556K]3 years ago
5 0

Answer:

it b

Step-by-step explanation:

4vir4ik [10]3 years ago
4 0

Answer:

Y=4X

Step-by-step explanation:

Hope this helps

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<h3>Need to Do :- </h3>
  • 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

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8 0
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