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find the dimensions of a rectangle that has a length 3 inches longer than it’s width and has an area of 40in^2

Respuesta :

fieryanswererft

Answer

length: 8 inch and width: 5 inch

Explanation

Let width be x,

then the length: x + 3

solve:

[tex]\sf (x+3)(x) = 40[/tex]

[tex]\sf x^2+3x = 40[/tex]

[tex]\sf x^2+3x-40=0[/tex]

[tex]\sf x^2+8x-5x-40=0[/tex]

[tex]\sf x(x+8)-5(x+8)=0[/tex]

[tex]\sf (x-5)(x+8)=0[/tex]

[tex]\sf x = 5 \ or \ -8[/tex]

  • As width cannot be negative, width will be 5.

Find for length:

[tex]\sf x + 3[/tex]

[tex]\sf 5+ 3[/tex]

[tex]\sf 8[/tex]

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