Find the first 4 terms of the sequence defined by this rule: f(n)=4n-7
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The area of polygon MNOPQR = Area of a rectangle that is 15 square units + Area of a rectangle that is 2 square units.
In the given polygon MNOPQR, side MN is parallel to side RQ and the side MR is parallel to side PQ
We will draw a perpendicular line from point O on the side RQ, which will intersect RQ at point S. So, we can now divide the whole polygon into two different rectangles MNSR and OPQS with the areas as A₁ and A₂ respectively.
In rectangle MNSR, length(MN) = 5 units and width (MR) = 3 units
According to the formula for Area of rectangle,
A₁ = (length)×(width)
A₁ = (5 units)×(3 units)
A₁ = 15 square units
Now in rectangle MNSR, side MN= side RS and side MR = side NS,
so RS= 5 units and NS= 3 units
That means, SQ= RQ- RS = 7-5 = 2 units
and OS= NS - NO = 3- 2 = 1 unit
In rectangle OPQS, we have length(SQ) = 2 units and width(OS) = 1 unit
So, A₂ = (length)×(width)
A₂ = (2 units)×(1 unit)
A₂ = 2 square units
So, the area of polygon MNOPQR = (Area of a rectangle that is 15 square units + Area of a rectangle that is 2 square units)
