You made a square card. The card did not fit in the envelope so you trimmed it. You trimmed 4 inches from the length and 5 inches from the width. The area of the resulting card is 20 square inches. What were the original dimensions for the card?

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.

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

left 3, up 4

step-by-step explanation:

just count the values of x & y of the vertex of the graph and see how many units in the x and y direction will get you from the f(x) graph to the f(x) graph

43

step-by-step explanation: