Question

find k for the following tables.

9

x	y
18.8	9.4
15.2	7.6
13	6.5
9	4.5
6.2	3.1

k = ______

10

x	y
\\(\frac{5}{6}\\)	\\(\frac{1}{2}\\)
1	\\(\frac{3}{5}\\)
\\(\frac{5}{3}\\)	1
\\(\frac{5}{2}\\)	\\(\frac{3}{2}\\)
10	6

k = ______

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Explanation:

Problem 9 (First Table)

Step1: Recall the formula for direct variation

For direct variation, \( y = kx \), so \( k=\frac{y}{x} \). We can use any pair of \( x \) and \( y \) values to find \( k \). Let's use the first pair \( x = 21 \), \( y = 10.5 \).
\( k=\frac{10.5}{21} \)

Step2: Calculate the value of \( k \)

\( \frac{10.5}{21}=0.5 \)
We can check with another pair, e.g., \( x = 6 \), \( y = 3 \) (wait, the last pair is \( x = 6.2 \), \( y = 3.1 \)). Let's check \( \frac{3.1}{6.2}=0.5 \). So \( k = 0.5 \).

Step1: Recall the formula for direct variation

For direct variation, \( y = kx \), so \( k=\frac{y}{x} \). Let's use the first pair \( x=\frac{1}{3} \), \( y=\frac{1}{5} \).
\( k=\frac{\frac{1}{5}}{\frac{1}{3}} \)

Step2: Simplify the fraction division

\( \frac{\frac{1}{5}}{\frac{1}{3}}=\frac{1}{5}\times\frac{3}{1}=\frac{3}{5}=0.6 \)
Let's check with another pair, e.g., \( x = 1 \), \( y=\frac{3}{5} \). \( \frac{\frac{3}{5}}{1}=\frac{3}{5}=0.6 \). Also, \( x = 10 \), \( y = 6 \), \( \frac{6}{10}=0.6 \). So \( k=\frac{3}{5} \) (or \( 0.6 \)).

Answer:

\( 0.5 \)

Problem 10 (Second Table)