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allochka39001 [22]

3 years ago

5

The length of a rectangle is x feet more than its width. The area of the rectangle is 5x + 25. What is the width of the rectangl

e?

1 answer:

AVprozaik [17]3 years ago

6 0

Answer:

w = $l + \frac{25 - l^{2}}{l - 5}$

(Anyone can correct me if I'm wrong)

Step-by-step explanation:

Before we start the solving, we can make the following statements:

Area = l x w

Area = 5x + 25

therefore,

l x w = 5x + 25

Since the question states that the length is x more than the width, so we can make the following statement:

w = l + x

With this, we can substitute it to the first statement we made, l x w = 5x + 25,

l x (l + x) = 5x + 25

$l^{2}$ + lx = 5x + 25

lx - 5x = 25 - $l^{2}$

x(l - 5) = 25 - $l^{2}$

x = $\frac{25 - l^{2}}{l - 5}$

From this, we can find w by substituting it in the statement we made earlier, w = l + x,

w = $l + \frac{25 - l^{2}}{l - 5}$

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Factor x^2+5x-50<br> A. (x+25)(x-2)<br> B. (x-5)(x+10)<br> C. (x-10)(x+5)<br> D. (x+2)(x-25)

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Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

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Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

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Substituting these equations in <em>c)</em>

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This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

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Remark

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