Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
Well if you are solving for y
the answer is
Y= 4/5x- 3
if you are solving for x
the answer is
X=5/4y + 15/4
*look at the image for the work*
I would think that all but one point would be on the line. One way to approach this problem is to find the equation of the line based upon any two points chosen at random, and then determine whether or not the other points satisfy this equation. Next time, would you please enclose the coordinates of each point inside parentheses: (2.5,14), (2.25,12), and so on, to avoid confusion.
14-12
slope of line thru 1st 2 points is m = ---------------- = 2/0.25 = 8
2.50-2.25
What is the eqn of the line: y = mx + b becomes
14 = (8)(2.5) + b; find b:
14-20 = b = -6. Then, y = 8x - 6.
Now determine whether (12,1.25) lies on this line.
Is 1.25 = 8(12) - 6? Is 1.25 = 90? No. So, unless I've made arithmetic mistakes, (1.25, 5) does not lie on the line thru (2.5,14) and (2.25,12).
Why not work this problem out yourself using my approach as a guide?
Answer:
74
Step-by-step explanation:
