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Liula [17]
3 years ago
14

The area of a rectangle is 256.5m2. if the length is 18m, what is the perimeter of the rectangle

Mathematics
2 answers:
Vlad [161]3 years ago
4 0
256.5/18= 14.25

(14.25x2)+(18x2)= 64.5m
lora16 [44]3 years ago
3 0

Perimeter of the rectangle is 64.5 m

<h3>Further explanation</h3>

To solve the above questions, we need to recall some of the formulas as follows:

Area of Rectangle = Length × Width

Perimeter of Rectangle = 2 × ( Length + Width )

Let us now tackle the problem !

<u>Given:</u>

Area of Rectangle = A = 256.5 m²

Length of Rectangle = L = 18 m

<u>Unknown:</u>

Perimeter of Rectangle = P = ?

<u>Solution:</u>

This problem is about Area and Perimeter of Rectangle.

<em>Let's find the width of Rectangle.</em>

\texttt{Area of Rectangle} = \texttt{Length} \times \texttt{Width}

256.5 = 18 \times \texttt{Width}

\texttt{Width} = 256.5 \div 18

\texttt{Width} = \boxed {14.25 ~ \texttt{m}}

<em>Let's find the perimeter of Rectangle.</em>

\texttt{Perimeter of Rectangle} = 2 (\texttt{Length} + \texttt{Width})

P = 2 ( L + W )

P = 2 ( 18 + 14.25 )

P = 2 ( 32.25 )

P = \boxed {64.5 ~ \texttt{m}}

<h3>Learn more</h3>
  • The perimeter of a polygon : brainly.com/question/6361596
  • The perimeter of a rectangle : brainly.com/question/7619923
  • The perimeter of a triangle : brainly.com/question/2299951

<h3>Answer details</h3>

Grade: College

Subject: Mathematics

Chapter: Two Dimensional Figures

Keywords: Perimeter, Area , Square , Rectangle , Side , Length , Width

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Step-by-step explanation:

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artcher [175]

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y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

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\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

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3 years ago
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