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$x= 0, 1, 2, 3, 4$ and $y= 26, 27, 18, 9, 0$ which of the following equations relates $y$ to $x$ for the values in the table?

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Prime factorization with exponents

Full solution

Q.

x= 0, 1, 2, 3, 4

and

y= 26, 27, 18, 9, 0

which of the following equations relates

y

to

x

for the values in the table?

Identify pattern in x-values: Identify the pattern in the x-values. The x-values are increasing by $1$ each time, starting from $0$ and going up to $4$ .
Identify pattern in $y$ -values: Identify the pattern in the $y$ -values. The $y$ -values are decreasing, but not in a linear way. We need to find a relationship that explains this decrease.
Check for constant difference: Check if the y-values are decreasing by a constant difference. Calculate the differences between consecutive y-values: $27 - 26 = 1$ , $18 - 27 = -9$ , $9 - 18 = -9$ , $0 - 9 = -9$ .
Find quadratic relationship: Notice that the differences are not constant, which suggests that the relationship between $x$ and $y$ is not linear. We need to look for a different type of relationship, possibly quadratic or exponential.
Consider exponential relationship: Try to find a quadratic relationship by checking if the differences of the differences are constant. Calculate the second differences: $(-9) - 1 = -10$ , $(-9) - (-9) = 0$ , $(-9) - (-9) = 0$ .
Determine constants $a$ and $b$ : Observe that the second differences are not constant, which suggests that the relationship is not quadratic. We need to consider other types of relationships.
Find value of $\newline$ $b$ : Consider an exponential relationship. Since the $\newline$ $y$ -values decrease as the $\newline$ $x$ -values increase, and the $\newline$ $y$ -value is $\newline$ $0$ when $\newline$ $x$ is $\newline$ $4$ , we can hypothesize that the relationship might be of the form $\newline$ $y = a \cdot b^x$ , where $\newline$ $b$ is a fraction (since the $\newline$ $y$ -values are decreasing) and $\newline$ $y$ equals $\newline$ $0$ when $\newline$ $x$ is $\newline$ $4$ .
Check if $\newline$ $b$ $\newline$ fits other points: Use the given $\newline$ $y$ $\newline$ -values to determine the constants $\newline$ $a$ $\newline$ and $\newline$ $b$ $\newline$ . When $\newline$ $x = 0$ $\newline$ , $\newline$ $y = 26$ $\newline$ , so we have $\newline$ $26 = a \cdot b^0$ $\newline$ . Since anything raised to the power of $\newline$ $0$ $\newline$ is $\newline$ $1$ $\newline$ , we have $\newline$ $a = 26$ $\newline$ .
Calculate math error: Now, use another point to find $b$ . When $x = 1$ , $y = 27$ , so we have $27 = 26 \cdot b^1$ . Solving for $b$ , we get $b = \frac{27}{26}$ .
Calculate math error: Now, use another point to find $b$ . When $x = 1$ , $y = 27$ , so we have $27 = 26 \cdot b^1$ . Solving for $b$ , we get $b = \frac{27}{26}$ .Check if the value of $b = \frac{27}{26}$ fits the other points in the table. When $x = 2$ , $y$ should be $18$ . Check if $x = 1$ $0$ . Calculate $x = 1$ $1$ and multiply by $x = 1$ $2$ .
Calculate math error: Now, use another point to find $b$ . When $x = 1$ , $y = 27$ , so we have $27 = 26 \cdot b^1$ . Solving for $b$ , we get $b = \frac{27}{26}$ .Check if the value of $b = \frac{27}{26}$ fits the other points in the table. When $x = 2$ , $y$ should be $18$ . Check if $x = 1$ $0$ . Calculate $x = 1$ $1$ and multiply by $x = 1$ $2$ .Calculate $x = 1$ $3$ . Then multiply by $x = 1$ $2$ to get $x = 1$ $5$ . This does not equal $18$ , so there is a math error in our hypothesis.

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