The area of the rectangular picture with a width 2 inches less than its length is 48 square inches. What are .? - picture of albertas operator license
The surface of the rectangular image with a width of 2 cm shorter than the length is 48 inches square. What are the dimensions of the image?
Friday, December 18, 2009
Picture Of Albertas Operator License The Area Of The Rectangular Picture With A Width 2 Inches Less Than Its Length Is 48 Square Inches. What Are .?
4 comments:
WW:
w (w + 2) = 48
² W + 2 W = 48
w ² + w ² = 48 + 1
w = w ² + 48 + 1
(W + 1) ² = 49
w + 1 = 7
w = 6
Length:
= 6 + 2
= 8
Answer: 6 cm to 8 cm, the dimensions.
May you give the width in terms of length, L is the length and width is two inches shorter than L-or R-2. The formula to calculate the area of a rectangle is A = LW. Therefore, replacing values that are in the formula:
A = LW A = 48 L = L, L = L-2
48 = L (L-2) distribution of L:
48 = L ^ 2-2L Now, subtract 48 from each side to a quadratic equation:
0 = L ^ 2-2L-48 and factors. L ^ 2 can be taken as L and L, so set in their brackets:
(L) (L). Now look at the last link, -48. As it seems to, such factors are a positive and a negative if they have your brackets: (+ L) (L-). Now, it is necessary, factor of -48 can be found, adding that if the average value of -2. 6 and -8 meet these requirements, we must
0 = (R 6) (L-8) If you multiply two numbers and then get a product of 0, at least one of these numbers should be 0. Since we do not know that every word is equal to 0 and solve for L:
L 6 = 0
L = -6 even though it is mathematically correct, you can not over a rectangle with a length that isNegative, contempt for the answer.
L-8 = 0
L = 8 Therefore, the length of the rectangle 8 inches, and use the original expression of the width should be noted that:
W = L-2, L = 8
w = 8.2
w = 6
Therefore, the rectangle is 8 cm long and 6 inches wide.
6 inches x 8 inches
6 inches x 8 inches
Post a Comment