Let the number of rectangles = R
Perimeter is 2(L) + 2(W)
The length remains the same when you add more triangles ( 6).
The width changes by multiplying the width by the number of rectangles.
Since the width of the rectangles are 1, multiply the number of rectangles R by 1 to get R.
Now replace W in the perimeter frmula with R:
Now the perimetr would be: P = 2L + 2R
Let's put this in terms of a system of equations. Let's call the numbers x and y, with x being the larger integer and y the smaller one. Since they are consecutive odd integers, we know that:
x = y + 2
The constraint is:
xy= 3(x+y) + 71 Simplify
xy = 3x + 3y + 71 Plug in (y+2) for x:
(y + 2) y = 3(y + 2) + 3y + 71 Simplify
y² + 2y = 3y + 6 + 3y + 71 Subtract 6y from both sides, add 71+6
y² -4y = 77
y² - 4y - 77 = 0
Factor:
(y - 11) (y+7) = 0
y = 11 or y=-7
When y = 11, x = 13, since x = y+2
Plug in to make sure:
xy = 3(x+y) + 71
11*13 = 3 (11 + 13) + 71
143 = 3 (24) + 71
143 = 72 + 71
143 = 143
Our values work!
Answer: 11, 13
Answer: 7.50 in paper and 35 in ink.
Step-by-step explanation:
