+1124 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.
+26393 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 \(y= \left(\dfrac{x+1}{2}\right)\times 5\)
+1124 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 \({(x+1)\over2}*5\)
+26393 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
\(\dots~ y = 5n\)
