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Zigmanuir [339]
3 years ago
6

Creat an equation for a cubic function, in standard form, that has x- intercept given by the set {-3,1,7} and which passes throu

gh the points (-2,54)
Mathematics
1 answer:
prohojiy [21]3 years ago
6 0

Given :

Set of x-intercepts {-3,1,7},

point through which it passes (-2,54)

Now,

Step 1: Substitute value of x intercepts in equation, y=a(x-p)(x-q)(x-r), we get,

y=a(x-(-3))(x-1)(x-7)

y= a(x+3)(x-1)(x-7)               ... equation (1)

Step 2: substitute x an y with the point through which it passes,

54=a((-2)+3)((-2)-1)((-2)-7)

54 = 27a

∴ a=2

Step 3: Now, substituting value of a in equation (1), we have

y=(2x+6)(x^{2} -8x+7)

y=2x^{3}-10x^{2}  -34x +42   (Requited cubic equation)

   

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WARNING : NON SENSE ANSWER REPORT TO MODERATOR ​
Rzqust [24]

Answer:

\frac{180}{147}

Step-by-step explanation:

  • Simplify (\frac{3}{-7} - \frac{11}{21} )

=> \frac{(-3 \times 3) - 11}{21}

=> \frac{-9 - 11}{21}

=> \frac{-20}{21}

  • Find the additive inverse of  \frac{-20}{21} by using its property - <em>"Sum of a number & its additive inverse is always zero". </em>Assume that 'x' is an additive inverse of  \frac{-20}{21}.

=> x + \frac{-20}{21} = 0

=> x = 0 - (-\frac{20}{21}) = \frac{20}{21}

  • Simplify (\frac{9}{5} \div \frac{7}{5} )

=> \frac{9}{5} \times \frac{1}{\frac{7}{5} }

=> \frac{9}{5} \times \frac{5}{7}

=> \frac{9}{7}

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=> \frac{9}{7} \times \frac{20}{21}

=> \frac{180}{147}

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