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avatar+846 

Hi friends, please help me with finding the rUle for these inputs aand outputs:

 

Inputs are 1; 3; 5 ;7 (X values)

Outputs are 5; 10; 15; 20 (Y values)

 

I did the following:

constant difference is according to the outputs, which is 5.

 the formule used is y=bx+c, b equal to the constant difference

 

so we have y=5x+c.

So if I take the first term for example, to calculate "c"

 

y=5(x)+c

5=5(1)+c

c=0

 

however, using c=0, does not calculate correctly for the other terms. also, "c" calculates differently for each term. Please help, I really do appreciate.

 May 23, 2021
 #1
avatar+25928 
+3

Hi friends, please help me with finding the rUle for these inputs aand outputs:

 

Inputs are 1; 3; 5 ;7 (X values)

Outputs are 5; 10; 15; 20 (Y values)

 

\(\begin{array}{|r|r|r|r|} \hline x & \dfrac{x+1}{2} & \left(\dfrac{x+1}{2}\right) \times 5 \\ \hline 1 & \dfrac{1+1}{2}= 1 & 1*5 = 5 \\ \hline 3 & \dfrac{3+1}{2}= 2 & 2*5 = 10\\ \hline 5 & \dfrac{5+1}{2}= 3 & 3*5 = 15\\ \hline 7 & \dfrac{7+1}{2}= 4 & 4*5 = 20 \\ \hline \end{array}\)


The rule is \(y= \left(\dfrac{x+1}{2}\right)\times 5\)

laugh

 May 23, 2021
 #2
avatar+846 
+1

Hi heureka,

 

Thank you for the response, however, could you kindly explain how you got to the rule in the first place...how did you know it had to be \({(x+1)\over2}*5\)

juriemagic  May 23, 2021
 #3
avatar+25928 
+2

how you got to the rule in the first place:

Arithmetic progression: \(1,~ 3,~ 5,~ 7,~\ldots\)
\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)*d \quad | \quad a_1 = 1,~ d=2 \\ a_n &=& 1+(n-1)*2 \\ a_n &=& 1+2n -2 \\ a_n &=& 2n-1 \\ 2n &=& a_n + 1 \\ n &=& \dfrac{a_n+1}{2} \quad \text{or} \quad n = \dfrac{x+1}{2}\\ \hline \end{array}\)

source: https://en.wikipedia.org/wiki/Arithmetic_progression

 

\(\dots~ y = 5n\)

 

laugh

heureka  May 23, 2021
edited by heureka  May 23, 2021
edited by heureka  May 23, 2021

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