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anonymous
 one year ago
Where is the removable discontinuity of f(x)=(x+5)/(2x^2+3x10) located
a.x=5
b.x = –2
c.x = 2
d.x = 5
anonymous
 one year ago
Where is the removable discontinuity of f(x)=(x+5)/(2x^2+3x10) located a.x=5 b.x = –2 c.x = 2 d.x = 5

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please can you check your formula of f(x)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0is the denominator, like this: \[2{x^2} + 3x  10\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the solution of this equation: \[2{x^2} + 3x  10 = 0\] are: \[\begin{gathered} x = \frac{{  3 \pm \sqrt {{3^2}  4 \times 2 \times \left( {  10} \right)} }}{{2 \times 2}} = \hfill \\ \hfill \\ = \frac{{  3 \pm \sqrt {9 + 80} }}{4} = \frac{{  3 \pm \sqrt {89} }}{4} \hfill \\ \end{gathered} \] am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I have applied the standard formula: \[x = \frac{{  b \pm \sqrt {{b^2}  4 \times a \times c} }}{{2 \times a}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I think that the denominator can be this: \[{x^2} + 3x  10\] is it possible?
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