Devorah is filling a pool with a hose. The volume, [tex]H[/tex], in liters, of water coming out of the hose in [tex]m[/tex] minutes is given by the function [tex]H(m)=17.4m[/tex]. However, it is a sunny day, and water is also evaporating from the pool. Therefore, the volume, [tex]V[/tex], in liters, of water in the pool [tex]m[/tex] minutes after Devorah started filling it is given by [tex]V(m)=17m[/tex].

Let [tex]E[/tex] be the volume of water, in liters, that has evaporated from the pool [tex]m[/tex] minutes after Devorah started filling it.

Write a formula for [tex]E(m)[/tex] in terms of [tex]H(m)[/tex] and [tex]V(m)[/tex].

Write a formula for [tex]E(m)[/tex] in terms of [tex]m[/tex].

Respuesta :

Answer:

Formula for [tex]E(m)[/tex] in terms of [tex]H(m)\ and\ V(m)[/tex] is:[tex]E(m)=H(m)-V(m)[/tex]

Formula for [tex]E(m)[/tex] in terms of [tex]m[/tex] is:

[tex]E(m)=0.4m[/tex]

Step-by-step explanation:

Given:

The expression for volume of water in a pool when there is no evaporation is:

[tex]H(m) =17.4m[/tex]

Where, [tex]m[/tex] is the minutes passed.

The expression for volume of water in a pool on a sunny day when there is evaporation of water is:

[tex]V(m)=17m[/tex]

Now, the change in the volume of the water in the pool is nothing but the volume of water that has been evaporated.

So, volume of water that has been evaporated is given as:

[tex]\textrm{Evaporated volume},x=H(m)-V(m)[/tex]

But as per question, evaporated volume of water is given by [tex]E(m)[/tex]. Therefore,

[tex]E(m)=H(m)-V(m)[/tex]

Now, plug in the values of [tex]H(m)\ and\ V(m)[/tex]

[tex]E(m)=17.4m-17m=(17.4-17)m=0.4m\\\therefore E(m)=0.4m[/tex]

Hence, formula for [tex]E(m)[/tex] in terms of [tex]H(m)\ and\ V(m)[/tex] is:

[tex]E(m)=H(m)-V(m)[/tex]

Formula for [tex]E(m)[/tex] in terms of [tex]m[/tex] is:

[tex]E(m)=0.4m[/tex]

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