Ask question

Ask question

All categories

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.

1 answer:

shutvik [7]1 year ago

5 0

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

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

You might be interested in

What is the speed of a car that traveled a<br> total of 300 kilometers north in 2 hours?

Vlad1618 [11]

Speed = Distance/Time

So, speed

= 300 km/2h

= 150km/h

= (150 × 1000m) / 3600s

= 150000m/3600s

= 41.66m/s [approximately]

5 0

3 years ago

How do I solve these????

spin [16.1K]

#10

15(1/3y-3/5x-2/3y+4/15x)

Calculate

15(-y/3-x/3)

Write all numerator over common denominator

15(-y+x/3)

Multiply by reducing

-5(y+x)

Multiply parenthesis by -5

-5y-5x

#11

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

Reduce fraction

24(2/3y-3/4x-1/4y+5/6x)

Calculate

24(5/12y+x/12)

Multiply parenthesis by 24

10y + 2x

7 0

3 years ago

What polygon does the new shape resemble?

Masteriza [31]

Answer:

Triangle

Step-by-step explanation:

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

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

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

6 0

3 years ago

Fabiola found that she has already used 60 pieces, which is 40% of her notebook paper. How many pieces of paper were in the note

jok3333 [9.3K]

I believe the correct answer would be 150 pages

4 0

3 years ago