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Dimas [21]
1 year ago
8

The CEO of NJAD, a Real Estate Investment Fund, decided to design a new development inPolycity, New Jersey. She wants to build 3

models as followsModel A has a swimming poolModel C is a parallelogram with parallel and congruent opposite sidesModel B is a trapezoid with a large backyarda) Find the perimeter of Model A above.

Mathematics
1 answer:
shutvik [7]1 year ago
5 0

The perimeter (P) is the sum of all sides:

\begin{gathered} (2x^2-6x+25)+(5x^3+6x^2-3x+4)+(x^3-2x^2+8x+9)+(3x^4+2x^3+8x+7) \\ 2x^2-6x+25+5x^3+6x^2-3x+4+x^3-2x^2+8x+9+3x^4+2x^3+8x+7 \\ 3x^4+5x^3+x^3+2x^3+2x^2+6x^2-2x^2-6x-3x+8x+8x+25+4+9+7 \\ 3x^4+8x^3+6x^2+7x+45 \end{gathered}

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Vlad1618 [11]

Speed = Distance/Time

So, speed

= 300 km/2h

= 150km/h

= (150 × 1000m) / 3600s

= 150000m/3600s

= 41.66m/s [approximately]

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How do I solve these????
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#10


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24(2/3y-3/4x-2/8y+5/6x)


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3 years ago
What polygon does the new shape resemble?
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NICE try :)

4 0
3 years ago
Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a
Crank

I'll assume the ODE is actually

y''+(x-2)y'+y=0

Look for a series solution centered at x=2, with

y=\displaystyle\sum_{n\ge0}c_n(x-2)^n

\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n

\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n

with c_0=y(2)=2 and c_1=y'(2)=0.

Substituting the series into the ODE gives

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0

\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}

  • If n=2k for integers k\ge0, then

k=0\implies n=0\implies c_0=c_0

k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}

k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}

k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}

and so on, with

c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}

  • If n=2k+1, we have c_{2k+1}=0 for all k\ge0 because c_1=0 causes every odd-indexed coefficient to vanish.

So we have

y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}

Recall that

e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}

The solution we found can then be written as

y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k

\implies\boxed{y(x)=2e^{-(x-2)^2/2}}

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