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2 years ago

The length of a rectangle is 4yd longer than its width. If the perimeter of the rectangle is 36yd, find its area

Mathematics

1 answer:

kicyunya [14]2 years ago

3 0

Answer:

$\boxed{\sf Area \ of \ the \ rectangle = 91 \ yd^{2}}$

Given:

Length of the rectangle = 4 yd longer than its width

Perimeter of the rectangle = 36 yd

To Find:

Area of the rectangle

Step-by-step explanation:

Let the width of the rectangle be 'w' yd

So,

Length of the rectangle = (w + 4) yd

$\therefore \\ \sf \implies Perimeter \: of \: the \: rectangle = 2(Length + Width) \\ \\ \sf \implies 36 = 2((4 + w) + w) \\ \\ \sf \implies 36 = 2(4 + w + w) \\ \\ \sf \implies 36 = 2(4 + 2w) \\ \\ \sf 36 =2(2w+4) \: is \: equivalent \: to \: 2(2w + 4) = 36: \\ \sf \implies 2(2w + 4) = 36 \\ \\ \sf Divide \: both \: sides \: of \: 2 (2w + 4) = 36 \: by \: 2: \\ \sf \implies 2w + 4 = 18 \\ \\ \sf Subtract \: 4 \: from \: both \: sides: \\ \sf \implies 2w = 14 \\ \\ \sf Divide \: both \: sides \: of \: 2w = 14 \: by \: 2: \\ \sf \implies w = 7$

So,

Width of the rectangle = 7 yd

Length of the rectangle = (7 + 4) yd

= 13 yd

$\therefore \\ \sf Area \ of \ the \ rectangle = Length \times Width \\ \\ \sf = 7 \times 13 \\ \\ \sf = 91 \: {yd}^{2}$

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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:

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