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3 years ago

Can someone help meeeeee

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

7 0

Answer:

Area of shaded rectangle = 42 cm²

55%

Step-by-step explanation:

The length of the shaded area is equal to the length of the white rectangle.

Therefore, length of shaded rectangle = 7 cm

The width of the shaded rectangle is equal to the length of the white rectangle <u>minus</u> two widths of the white rectangle.

Therefore, width = 7 - (2 x 0.5) = 6 cm

Area of shaded rectangle = width x length

= 6 x 7

= 42 cm²

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Perimeter of a square = 4 × side length

If the perimeter of the square is 40 cm,
then the side length = 40 ÷ 4 = 10 cm

Area of a square = side length x side length

⇒ area of this square = 10 x 10 = 100 cm²

To determine the shaded area, calculate the areas of the 2 white triangles and subtract these from the area of the square.

Area of a triangle = 1/2 x base x height

⇒ area of left triangle = 1/2 x (10 - 7) x 10 = 15 cm²

⇒ area of right triangle = 1/2 x 10 x (10 - 4) = 30 cm²

Therefore, shaded area = 100 - 15 - 30 = 55 cm²

To calculate the percentage, divide the shaded area by the area of the square and multiply by 100:

(55 ÷ 100) × 100% = 55%

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Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$