2l+2w=perimeter <= equation for perimeter
w+4=l <=length is 4 greater than width 
2w+2(w+4)=24 <= plug in the l in the first equation with the second equation 
2w+2w+8=24 <= solve the equation 
4w=16 
w=4 
and, since 
w+4=l <= length is 4 greater than width 
then l=4+4 
l=8

4 0

4 years ago

Read 2 more answers

Please Help me!!! See picture in attachments

liubo4ka [24]

Answer:

The answer would be 15.

Step-by-step explanation:

2x + 7 = x +4

2x = . x +1

x =11

2(11) =7

22-7

15.

5 0

3 years ago

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Solve the given initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1

Yuri [45]

Let's check if the ODE is exact. To do that, we want to show that if

$\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0$

then $M_y=N_x$ . We have

$M_y=2(x+y)$
$N_x=2y+2x=2(x+y)$

so the equation is indeed exact. We're looking for a solution of the form $\Psi(x,y)=C$ . Computing the total differential yields the original ODE,

$\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$
$\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}$

Integrate both sides of the first PDE with respect to $x$ ; then

$\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)$

where $f(y)$ is a function of $y$ alone. Differentiate this with respect to $y$ so that

$\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)$
$\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)$
$f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C$

So the solution to this ODE is

$\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C$

i.e.

$\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C$

6 0

3 years ago