Question

practice & problem solving
practice
do the tables of values represent inverse variations? explain. see example 1
14.

x	-\\(\frac{1}{4}\\)	-\\(\frac{1}{2}\\)	\\(\frac{1}{3}\\)	2	5	11
y	-\\(\frac{9}{2}\\)	-9	6	36	90	198

15.

x	1	2	3	4	5	6
y	60	30	20	15	12	10

1. if x and y vary inversely and x = 3 when y = \\(\frac{3}{2}\\), what is the value of y when x = -1?

see example 2

Explanation:

Problem 14:

Step1: Recall inverse variation formula

For inverse variation, \( xy = k \) (constant) for all pairs \((x, y)\).

Step2: Calculate \( xy \) for each pair

• For \( x = -\frac{1}{4}, y = -\frac{9}{4} \): \( (-\frac{1}{4})(-\frac{9}{4}) = \frac{9}{16} \)
• For \( x = -\frac{1}{2}, y = -9 \): \( (-\frac{1}{2})(-9) = \frac{9}{2} \)
• For \( x = \frac{1}{3}, y = 6 \): \( (\frac{1}{3})(6) = 2 \)
• For \( x = 2, y = 36 \): \( (2)(36) = 72 \)
• For \( x = 5, y = 90 \): \( (5)(90) = 450 \)
• For \( x = 11, y = 198 \): \( (11)(198) = 2178 \)

Since \( xy \) is not constant, the table does not represent inverse variation.

Step1: Recall inverse variation formula

For inverse variation, \( xy = k \) (constant) for all pairs \((x, y)\).

Step2: Calculate \( xy \) for each pair

• For \( x = 1, y = 60 \): \( (1)(60) = 60 \)
• For \( x = 2, y = 30 \): \( (2)(30) = 60 \)
• For \( x = 3, y = 20 \): \( (3)(20) = 60 \)
• For \( x = 4, y = 15 \): \( (4)(15) = 60 \)
• For \( x = 5, y = 12 \): \( (5)(12) = 60 \)
• For \( x = 6, y = 10 \): \( (6)(10) = 60 \)

Since \( xy = 60 \) (constant) for all pairs, the table represents inverse variation.

Step1: Recall inverse variation formula

Inverse variation: \( xy = k \). Given \( x = 3, y = \frac{3}{2} \), find \( k \).
\( k = (3)(\frac{3}{2}) = \frac{9}{2} \)

Step2: Find \( y \) when \( x = -1 \)

Using \( xy = k \), substitute \( x = -1 \) and \( k = \frac{9}{2} \):
\( (-1)y = \frac{9}{2} \)
\( y = -\frac{9}{2} \)

Answer:

No, because \( xy \) is not constant for all pairs.

Problem 15: