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joja [24]
3 years ago
11

If the first of three consecutive integers is subtracted from 138, the result is the sum of the second and third. What are the i

ntegers?
Mathematics
1 answer:
otez555 [7]3 years ago
8 0

Answer:

First Integer = n = 45

Second Integer = n+1 = 45 + 1 = 46

And Third Integer = n+ 2 = 45 +2 = 47

Step-by-step explanation:

Let First integer = n

Second Integer = n+1

Third Integer = n+2

According to the question given (If the first of three consecutive integers is subtracted from 138, the result is the sum of the second and third) the equation will be:

138 - n = (n+1) + (n+2)

Solving the equation:

138 - n = n+1+n+2

138 - n = 2n+3

138 - 3 =2n +n

135 = 3n

135/3 = n

=> n= 45

So, First Integer = n = 45

Second Integer = n+1 = 45 + 1 = 46

And Third Integer = n+ 2 = 45 +2 = 47

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irinina [24]

Answer:

\displaystyle  \frac{  {3x}^{2} }{ 35 }  +  \frac{{2y}^{2} }{  35  }   = 1

Step-by-step explanation:

we want to figure out the ellipse equation which passes through <u>(</u><u>1</u><u>,</u><u>4</u><u>)</u><u> </u>and <u>(</u><u>-</u><u>3</u><u>,</u><u>2</u><u>)</u>

the standard form of ellipse equation is given by:

\displaystyle  \frac{(x - h {)}^{2} }{ {a}^{2} }  +  \frac{(y - k {)}^{2} }{ {b}^{2} }  = 1

where:

  • (h,k) is the centre
  • a is the horizontal redius
  • b is the vertical radius

since the centre of the equation is not mentioned, we'd assume it (0,0) therefore our equation will be:

\displaystyle  \frac{  {x}^{2} }{ {a}^{2} }  +  \frac{{y}^{2} }{ {b}^{2} }  = 1

substituting the value of x and y from the point (1,4),we'd acquire:

\displaystyle  \frac{ 1}{ {a}^{2} }  +  \frac{16}{ {b}^{2} }  = 1

similarly using the point (-3,2), we'd obtain:

\displaystyle  \frac{ 9}{ {a}^{2} }  +  \frac{4 }{ {b}^{2} }  = 1

let 1/a² and 1/b² be q and p respectively and transform the equation:

\displaystyle  \begin{cases} q  +  16p  = 1  \\ 9q + 4p = 1 \end{cases}

solving the system of linear equation will yield:

\displaystyle  \begin{cases} q   =  \dfrac{3}{35} \\ \\  p =  \dfrac{2}{35}  \end{cases}

substitute back:

\displaystyle  \begin{cases}  \dfrac{1}{ {a}^{2} }   =  \dfrac{3}{35} \\ \\   \dfrac{1}{ {b}^{2} }  =  \dfrac{2}{35}  \end{cases}

divide both equation by 1 which yields:

\displaystyle  \begin{cases}  {a}^{2}   =  \dfrac{35}{ 3} \\ \\    {b}^{2}   =  \dfrac{35}{2}  \end{cases}

substitute the value of a² and b² in the ellipse equation , thus:

\displaystyle  \frac{  {x}^{2} }{  \dfrac{35}{3}  }  +  \frac{{y}^{2} }{  \dfrac{35}{2}  }   = 1

simplify complex fraction:

\displaystyle  \frac{  {3x}^{2} }{ 35 }  +  \frac{{2y}^{2} }{  35  }   = 1

and we're done!

(refer the attachment as well)

8 0
3 years ago
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