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Goshia [24]
3 years ago
10

Solve for the 5th term f(n)=3n/2n^2

Mathematics
1 answer:
kipiarov [429]3 years ago
3 0
F(5) = 5(5)/2(5)^2 = 25/2(25) = 25/50 = 1/2
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A rectangle with a length of L and a width of W has a diagonal of 10 inches. Express the perimeter P of the rectangle as a funct
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<h2>Answer:</h2>

The expression which represents the perimeter P of the rectangle as a function of L is:

          Perimeter=2(L+\sqrt{100-L^2})

<h2>Step-by-step explanation:</h2>

The length and width of a rectangle are denoted by L and W respectively.

Also the diagonal of a rectangle is: 10 inches.

We know that the diagonal of a rectangle in terms of L and W are given by:

10=\sqrt{L^2+W^2}

( Since, the diagonal of a rectangle act as a hypotenuse of the right angled triangle and we use the Pythagorean Theorem )

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10^2=L^2+W^2\\\\i.e.\\\\W^2=100-L^2\\\\W=\pm \sqrt{100-L^2}

But we know that width can't be negative. It has to be greater than 0.

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