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The number of cars (c) in a parking lot increases when the parking lot fee (f) decreases. Right there correct equation for the scenario, and for the number of cars when the fee is six dollars.

The number of cars c in a parking lot increases when the parking lot fee f decreases Right there correct equation for the scenario and for the number of cars wh class=

Respuesta :

If fee be y and cars be x

[tex]\\ \bull\tt\longmapsto x\propto \dfrac{1}{y}[/tex]

[tex]\\ \bull\tt\longmapsto x_1y_1=x_2y_2[/tex]

[tex]\\ \bull\tt\longmapsto 30(10)=x_2(6)[/tex]

[tex]\\ \bull\tt\longmapsto 300=6x_2[/tex]

[tex]\\ \bull\tt\longmapsto x_2=\dfrac{300}{6}[/tex]

[tex]\\ \bull\tt\longmapsto x_2=50[/tex]

Answer:

This is an inverse proportionality

[tex]{ \tt{c \: \alpha \: - \frac{1}{f} }} \\ \\ { \tt{c = - \frac{k}{f} }}[/tex]

• k is a constant of proportionality

• when c is 15, f is 20, therefore;

[tex]{ \tt{15 = \frac{ - k}{20} }} \\ \\ { \tt{ - k = 20 \times 15}} \\ \\ { \tt{k = - 300}}[/tex]

• Therefore, the equation is :

[tex]{ \boxed{ \bf{c = \frac{ - 300}{f} }}}[/tex]

→ when f is 10:

[tex]{ \tt{c = \frac{ - 300}{10} }} \\ \\ { \tt{c = - 30}}[/tex]

negative shows decrease

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