Question

name: cfa bse - 8th grade math
you must show work to support all of your answers throughout this cfa to receive full credit.

1. circle any graphs or tables below that represent a direct - variation relationship.
2. the value of y varies directly with x. when y is 15, x is 3. use this information to complete the table.

a.

x	0	1	2	3	4	5
y

b. what is the value of y when x is 2 1/2?
c. explain how you found your answer.

Explanation:

Step1: Recall direct - variation formula

The formula for direct variation is $y = kx$, where $k$ is the constant of variation. For a graph, a direct - variation is a straight line passing through the origin $(0,0)$. For a table, $\frac{y}{x}$ should be a constant for all non - zero $x$ values.

Step2: Analyze graphs

The first graph is a straight line but does not pass through the origin, so it is not a direct variation. The second graph is a straight line passing through the origin, so it represents a direct variation. The third graph is not a straight line, so it is not a direct variation.

Step3: Analyze tables

For the first table: $\frac{2}{3}
eq\frac{4}{6}$, so it is not a direct variation. For the second table: $\frac{4}{6}=\frac{8}{12}=\frac{12}{18}=\frac{16}{24}=\frac{20}{30}=\frac{2}{3}$, so it represents a direct variation. For the third table: $\frac{17}{2}
eq\frac{31}{4}$, so it is not a direct variation.

Step4: For part 2a

Given $y = kx$ and when $y = 15$, $x = 3$. First, find $k$: $k=\frac{y}{x}=\frac{15}{3}=5$. Then when $x = 0$, $y=k\times0 = 0$; when $x = 1$, $y = 5\times1=5$; when $x = 2$, $y = 5\times2 = 10$; when $x = 3$, $y = 15$; when $x = 4$, $y = 5\times4=20$; when $x = 5$, $y = 5\times5 = 25$.

Step5: For part 2b

If $x = 2\frac{1}{2}=\frac{5}{2}$ and $k = 5$, then $y=k\times x=5\times\frac{5}{2}=\frac{25}{2}=12.5$.

Step6: For part 2c

We first found the constant of variation $k$ using the given values $y = 15$ and $x = 3$ ($k=\frac{y}{x}$). Then we used the direct - variation formula $y = kx$ to find the values of $y$ for different $x$ values.

Answer:

1. Circle the second graph and the second table.

2a.

$x$	$y$
1	5
2	10
3	15
4	20
5	25

2b. $12.5$
2c. First, found $k=\frac{y}{x}$ using given $y = 15$ and $x = 3$. Then used $y = kx$ to find other $y$ values.