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
Answer:
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Step-by-step explanation:
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Hope this helps
Assuming BOD and AOC are straight lines.
Area of the rectangle = 4 x ΔAOD
Area of the rectangle = 4 x 10
Area of the rectangle = 40 unit²
Area of ΔABC = 1/2 x 40
Area of ΔABC = 20 unit²
Answer:
11 laps i think
Step-by-step explanation:
mark brainliest and have a great day!
Answer:
see explanation
Step-by-step explanation:
These are the terms of an arithmetic sequence with n th term
= a + (n - 1)d
where a is the first term and d the common difference
d = 25 - 20 = 30 - 25 = 5 and a = 20, hence
= 20 + 5(n - 1) = 20 + 5n - 5 = 5n + 15 ← n th term formula
