Answer:
(a) [tex]y = \frac{4}{x^2}[/tex]
(b) [tex]x = \frac{2}{5}[/tex]
Step-by-step explanation:
Given
Variation: Inverse proportional.
This is represented as:
[tex]y\ \alpha\ \frac{1}{x^2}[/tex]
See attachment for table
Solving (a):
First convert variation to equation
[tex]y = k\frac{1}{x^2}[/tex]
From the table:
[tex](x,y) = (1,4)[/tex]
So, we have:
[tex]4 = k * \frac{1}{1^2}[/tex]
[tex]4 = k * \frac{1}{1}[/tex]
[tex]4 = k * 1[/tex]
[tex]4 = k[/tex]
[tex]k = 4[/tex]
Substitute 4 for k in [tex]y = k\frac{1}{x^2}[/tex]
[tex]y = 4 * \frac{1}{x^2}[/tex]
[tex]y = \frac{4}{x^2}[/tex]
Solving (b): x when y = 25.
Substitute 25 for y in [tex]y = \frac{4}{x^2}[/tex]
[tex]25 = \frac{4}{x^2}[/tex]
Cross Multiply
[tex]25 * x^2 = 4[/tex]
Divide through by 25
[tex]x^2 = \frac{4}{25}[/tex]
Take positive square roots of both sides
[tex]x = \sqrt{\frac{4}{25}[/tex]
[tex]x = \frac{2}{5}[/tex]
