Respuesta :
Answer:
The original dimensions of the card were 9 inches length and 9 inches width
Step-by-step explanation:
Let x be the original length and width of the card in inches (remember that it was squared originally). The exercise says that the area of the new width, with dimensions x-4 and x-5 is 20 square inches, therefore
(x-4)*(x-5) = 20
x²-9x+20 = 20
x²-9x = 0
x*(x-9) = 0
x = 0 or x = 9
Since x must be positive, then it cant be 0, thus, x has to be 9. The original dimensions of the card were 9 inches length and 9 inches width.
Answer:
original dimension for the card = 9 inches by 9 inches
Step-by-step explanation:
Let the length and width of the card be x and x respectively since its a square card and they are both the same length.
Upon trimming, new length and width are (x-4) inches and (x-5) inches respectively.
Area of new card = (x-4)(x-5) = 20
Expanding the brackets to form a quadratic equation
Area of new card = x² - 5x - 4x + 20 = 20 ≡ x² -9x = 0
Hence x² = 9x
and x = 9
Therefore original dimension for the card = 9 inches by 9 inches