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ghari112345
01.07.2019 •
Mathematics
Find the dimensions of the box described. the length is twice as long as the width. the height is 4 inches greater than the width. the volume is 48 cubic inches. find the length, width, height
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Ответ:
Width of box= 2inches
Length of box= 4inches
Height of box= 6inches.
Step-by-step explanation:
Let width of box=x inches
Length of box = twice of width=
=![2x](/tpl/images/0040/0041/4bbd1.png)
Height of box= 4 inches greater than width=![x+4](/tpl/images/0040/0041/4b0d1.png)
Volume of box= 48 cubic inches
We know that the formula of volume of cuboid=![length\times breadth\times height](/tpl/images/0040/0041/75ebd.png)
Apply the formula
Volume of box=![x\times 2x\times (x+4)](/tpl/images/0040/0041/dd721.png)
Volume of cube =![2x^2(x+4)](/tpl/images/0040/0041/6c428.png)
Apply inspection method to solve the equation
Put![x=0](/tpl/images/0040/0041/37eb6.png)
Then we get![-24\neq0](/tpl/images/0040/0041/a34c3.png)
Hence, x=0 is not the solution of x
Put x=1 in the equation then we get
Hence x=1 is not the solution of equation.
Put x=2 then we get
8+16-24=0
Hence, x=2 is the solution of equation .
Now substitute equation
=0
Sum roots =6
Product of roots=12
When sum of roots is greater than zero and product of roots is greater than zero then value of roots of equation is imaginary.
Hence, the roots of equation
are imaginary.
Lenght , widht and height are dimensions of box therefore, imaginary value are not possible.
Hence,
is the only real values of root of equation .Therefore, it is possible and other two imaginary value of roots are not possible .
Widht of box=2 inches
Length of box =
=4inches
Height of box=
=2+4=6 inches
Ответ: