Write an equation that expresses the following relationship. d varies directly with the cube of w and inversely with p. In your equation, use k as the constant of proportionality.

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Riia

Here d varies directly with cube of w, and inversely with p. So w cube have to be in the numerator and p have to be in the denominator. And we need to use k for constant of proportionality. Therefore the required equation is

[tex] d=\frac{kw^3}{p} [/tex]

In direct proportionality, the relatioship is linear that is as w increases, d increases and as w decreases d decreases and for inverse, it is opposite .

The formula for the indicated variable is d = k w³ / p

Further explanation

Solving linear equation mean calculating the unknown variable from the equation.

Let the linear equation : y = mx + c

If we draw the above equation on Cartesian Coordinates , it will be a straight line with :

m → gradient of the line

( 0 , c ) → y - intercept

Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :

[tex]\large {\boxed {m = \frac{y_2 - y_1}{x_2 - x_1} } }[/tex]

If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :

[tex]\large { \boxed {y - y_1 = m ( x - x_1 ) } }[/tex]

Let us tackle the problem.

This problem is about directly and inversely proportional.

Given :

d varies directly with the cube of w → [tex]\boxed {d \propto w^3}[/tex]

d varies inversely with p → [tex]\boxed {d \propto p^{-1} }[/tex]

∴ [tex]\large {\boxed {d \propto ( w^3 \times p^{-1} )} }[/tex]

From the above relationship, we can write the equation for the variable d i.e:

[tex]\large { \boxed {d = k \frac{w^3}{p} } }[/tex]

Learn more

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Linear Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point

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