$y$ $y$ varies inversely with $x$ $x$ , and $x = 4.5$ $x=4.5$ when $y = 2.4$ $y=2.4$ . What is the value of $x$ $x$ when the value of $y = 4.32$ $y=4.32$ ?

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$y$ $y$ varies inversely with $x$ $x$ , and $x = 4.5$ $x=4.5$ when $y = 2.4$ $y=2.4$ . What is the value of $x$ $x$ when the value of $y = 4.32$ $y=4.32$ ?

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Answer 1 · 2018-06-06T15:19:50

$x = 2.5$ $color(blue)(x=2.5)$

Explanation:

Inverse variation is given by:

$y \propto \frac{k}{x^{n}}$ $y prop k/x^n$

Where $k$ $bbk$ is the constant of variation.

To find $k$ $bbk$ we substitute $x = 4.5$ $x=4.5$ and $y = 2.4$ $y=2.4$

$2.4 = \frac{k}{4.5}$ $2.4 = k/4.5$

$k = 2.4 \cdot 4.5 = 10.8$ $k=2.4*4.5=10.8$

When $y = 4.32$ $y=4.32$

$4.32 = \frac{10.8}{x}$ $4.32=10.8/x$

$x = \frac{10.8}{4.32} = 2.5$ $x=10.8/4.32=2.5$

Answer 2 · 2018-06-06T15:27:38

$x = 2.5$ $x=2.5$

Explanation:

Direct variation uses the equation $y = k x$ $y=kx$

Inverse variation uses the equation $y = \frac{k}{x}$ $y = k/x$

Where $k$ $k$ represents the constant of variation.

In order to solve this problem we need to use the numbers provided for the first scenario to solve for the constant of
variation $k$ $k$ and then use $k$ $k$ to solve for the second set of numbers.

$x_{1} = 4.5$ $x_1=4.5$
$y_{1} = 2.4$ $y_1=2.4$
$k = ?$ $k=?$

$y = \frac{k}{x}$ $y=k/x$ Equation of Inverse Variation

$2.4 = \frac{k}{4.5}$ $2.4 = k/4.5$
Use the multiplicative inverse to isolate $k$ $k$

$4.5*2.4 = k/cancel(4.5) * cancel4.5$

$k = 10.8$

$x_2=?$
$y_2=4.32$
$k=10.8$

$y=k/x$ Equation of Inverse Variation

$4.32 = 10.8/x$
Use the multiplicative inverse to bring $x$ out of the denominator

$x*4.32 = 10.8/cancel(x) * cancelx$

$4.32x = 10.8$

Divide both sides by $4.32$ to isolate $x$

$(cancel(4.32)x)/cancel4.32 = 10.8/4.32$

$x=2.5$

Answer 3 · 2018-06-06T15:28:03

$x = 2.5$

Explanation:

$y prop 1/x -> "Inverse variation"$

$y = k/x$ , where $k$ is constant

When;

$x = 4.5 and y = 2.4$

Substituting the values of $x and y$ into the equation..

$2.4 = k/4.5$

$2.4/1 = k/4.5$

Cross multiplying;

$2.4 xx 4.5 = k xx 1$

$10.8 = k$

Therefore;

$k = 10.8$

Now the relationship between the two unknowns becomes;

$y = 10.8/x$

What is $x$ when $y = 4.32$

Substituting the value of $y$ into the relationship equation..

$4.32 = 10.8/x$

$4.32/1 = 10.8/x$

Cross multiplying;

$4.32 xx x = 10.8 xx 1$

$4.32x = 10.8$

$x = 10.8/4.32$

$x = 2.5$

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$y$ $y$ varies inversely with $x$ $x$ , and $x = 4.5$ $x=4.5$ when $y = 2.4$ $y=2.4$ . What is the value of $x$ $x$ when the value of $y = 4.32$ $y=4.32$ ?

3 Answers

Explanation:

Explanation:

Explanation:

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yyy varies inversely with xxx, and x=4.5x=4.5x=4.5 when y=2.4y=2.4y=2.4. What is the value of xxx when the value of y=4.32y=4.32y=4.32?

3 Answers

Explanation:

Explanation:

Explanation:

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$y$ $y$ varies inversely with $x$ $x$ , and $x = 4.5$ $x=4.5$ when $y = 2.4$ $y=2.4$ . What is the value of $x$ $x$ when the value of $y = 4.32$ $y=4.32$ ?