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3 years ago

The length of a rectangular carpet is 3 feet longer than twice its width. If the area of

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

$Perimeter = 45ft$

Step-by-step explanation:

Given

Represent the Length as L and the Width as W.

So, we have the following:

$L = 2W + 3$

$Area = 65$

Required

Determine the perimeter

The area of a rectangle is calculated as thus:

$Area = L * W$

Substititute 65 for Area

$65 = L * W$

Substitute 2W + 3 for L

$65 = (2W + 3) * W$

This gives:

$65 = 2W^2 + 3W$

Equate to 0

$2W^2 + 3W - 65 = 0$

Expand:

$2W^2 + 10W - 13W - 65 = 0$

Factorize:

$2W(W + 5) - 13(W + 5) = 0$

$(2W - 13)(W + 5) = 0$

$2W - 13 = 0\ or\ W + 5 = 0$

$2W = 13\ or\ W = -5$

$W = \frac{13}{2}$ or $W = -5$

But width can't be less than 0.

So:

$W = \frac{13}{2}$

$W = 6.5$

Recall that:

$L = 2W + 3$

$L = 2 * 6.5 + 3$

$L = 16$

The perimeter is calculated as thus:

$Perimeter = 2 * (L + W)$

$Perimeter = 2 * (16 + 6.5)$

$Perimeter = 45ft$

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The equation given in the question has to be answered to the nearest tenth. The only thing that one should be careful about is the cross multiplication part. Other than that there is no difficulty in the given problem. Now let us get down to the equation in the question.
15/30 = n/34
30 * n = 15 * 34
30n = 510
n = 510/30
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Step-by-step explanation:

(8,8) is a solution is y=x.

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(8,8) is in (x,y) format

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3 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

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3 years ago