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How much water do I need to addto l liters of pure alcohol to obtain asolution of 45% alcohol? write in an equation form

How Much Water Do I Need To Addto L Liters Of Pure Alcohol To Obtain Asolution Of 45 Alcohol Write In An Equation Form class=

Sagot :

The perimeter of the rectangle is

[tex]=24[/tex]

The length is

[tex]=y[/tex]

The width is

[tex]=x[/tex]

The formula to calculate the perimeter of a rectangle is given as

[tex]=2(\text{length}+\text{width)}[/tex]

By substituting the values of the length, width, and perimeter above we will get the equation below

[tex]\begin{gathered} perimeter=2(\text{length}+\text{width)} \\ 2(y+x)=24 \end{gathered}[/tex]

To talk about the length y in terms of x, we will make y the subject of the formula in the equation above

[tex]\begin{gathered} 2(y+x)=24 \\ \text{divide both sides by 2} \\ \frac{2(y+x)}{2}=\frac{24}{2} \\ y+x=12 \\ \text{therefore,} \\ y=12-x \end{gathered}[/tex]

Hence,

The length y in terms of x is given as y=12-x

To calculate the possible values of x and the possible values of y

The sum of x and y must give 12 therefore,

[tex]\begin{gathered} \text{when x= 1} \\ y=12-x \\ y=12-1=11 \\ x=1,y=11 \end{gathered}[/tex][tex]\begin{gathered} \text{when x=2} \\ y=12-x \\ y=12-2 \\ y=10 \end{gathered}[/tex][tex]\begin{gathered} \text{when x=3} \\ y=12-x \\ y=12-3 \\ y=9 \end{gathered}[/tex][tex]\begin{gathered} \text{when x=4} \\ y=12-x \\ y=12-4 \\ y=8 \end{gathered}[/tex][tex]undefined[/tex]

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