Light shines through window what is the area of a trapezoid shaped region created by the light

Mathematics

1 answer:

bogdanovich [222]3 years ago

5 0

Answer:

16 ft²

Step-by-step explanation:

The complete question is attached.

A trapezoid is a quadrilateral (has four sides) with one a parallel base. The base angles and the diagonals of an isosceles trapezoid are equal.

The area of a trapezoid = [(sum of the parallel bases) / 2] * height of the trapezoid.

Given that the parallel bases are 3 ft and 5 ft, while the height of the trapezoid is 4 ft. Hence:

The area of a trapezoid = [(3 + 5)/2] * 4

The area of a trapezoid = 16 ft²

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Integration of ∫(cos3x+3sinx)dx

Murljashka [212]

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$\boxed{\pink{\tt I = \dfrac{1}{3}sin(3x) - 3cos(x) + C}}$

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

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Now , Rewrite using du and u .

$\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I = \dfrac{1}{3}sin(3x) - 3cos(x) + C }}}}}$