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3 years ago

X is inversely proportional to y, and x = 18 when y = 4. Write an equation relating the two

Mathematics

1 answer:

Alenkasestr [34]3 years ago

8 0

Answer:

$xy=72$

Step-by-step explanation:

If x is inversely porportionaly to y, then for some constant k,

$x = \frac{k}{y}\\xy=k$

Since y is 4 when x = 18, k is equal to 4*18=72, so the equation is:

$xy=72$

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What is the difference? StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction Sta

katovenus [111]

Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

$\frac{x}{x^2-16}-\frac{3}{x-4}$

To solve the given expression

$\frac{x}{x^2-16}-\frac{3}{x-4}$

$=\frac{x}{x^2-4^2}-\frac{3}{x-4}$

$=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}$ ( using the property $a^2-b^2=(a+b)(a-b)$ )

$=\frac{x-3(x+4)}{(x+4)(x-4)}$

$=\frac{x-3x-12}{(x+4)(x-4)}$ ( by using distributive property )

$=\frac{-2x-12}{(x+4)(x-4)}$

$=\frac{-2(x+6)}{(x+4)(x-4)}$

$\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

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