Subtract: (11w^3 + 15)

If an object is dropped from a height of 200 feet, the function h(t)=-16t^2=200 gives the height of the object after t seconds.

Fed [463]

Hello,
Answer C
h(t)=0=-16t²+200==>t=3.53533...(s)

5 0

3 years ago

Determine the values of the constants r and s such that i(x, y) = x rys is an integrating factor for the given differential equa

garri49 [273]

$\underbrace{y(7xy^2+6)}_{M(x,y)}\,\mathrm dx+\underbrace{x(xy^2-1)}_{N(x,y)}\,\mathrm dy=0$

For the ODE to be exact, we require that $M_y=N_x$ , which we'll verify is not the case here.

$M_y=21xy^2+6$
$N_x=2xy^2-1$

So we distribute an integrating factor $i(x,y)$ across both sides of the ODE to get

$iM\,\mathrm dx+iN\,\mathrm dy=0$

Now for the ODE to be exact, we require $(iM)_y=(iN)_x$ , which in turn means

$i_yM+iM_y=i_xN+iN_x\implies i(M_y-N_x)=i_xN-i_yM$

Suppose $i(x,y)=x^ry^s$ . Then substituting everything into the PDE above, we have

$x^ry^s(19xy^2+7)=rx^{r-1}y^s(x^2y^2-x)-sx^ry^{s-1}(7xy^3+6y)$
$19x^{r+1}y^{s+2}+7x^ry^s=rx^{r+1}y^{s+2}-rx^ry^s-7sx^{r+1}y^{s+2}-6sx^ry^s$
$19x^{r+1}y^{s+2}+7x^ry^s=(r-7s)x^{r+1}y^{s+2}-(r+6s)x^ry^s$
$\implies\begin{cases}r-7s=19\\r+6s=-7\end{cases}\implies r=5,s=-2$

so that our integrating factor is $i(x,y)=x^5y^{-2}$ . Our ODE is now

$(7x^6y+6x^5y^{-1})\,\mathrm dx+(x^7-x^6y^{-2})\,\mathrm dy=0$

Renaming $M(x,y)$ and $N(x,y)$ to our current coefficients, we end up with partial derivatives

$M_y=7x^6-6x^5y^{-2}$
$N_x=7x^6-6x^5y^{-2}$

as desired, so our new ODE is indeed exact.

Next, we're looking for a solution of the form $\Psi(x,y)=C$ . By the chain rule, we have

$\Psi_x=7x^6y+6x^5y^{-1}\implies\Psi=x^7y+x^6y^{-1}+f(y)$

Differentiating with respect to $y$ yields

$\Psi_y=x^7-x^6y^{-2}=x^7-x^6y^{-2}+\dfrac{\mathrm df}{\mathrm dy}$
$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

Thus the solution to the ODE is

$\Psi(x,y)=x^7y+x^6y^{-1}=C$

4 0

3 years ago

Why does fast- moving grader erosion and then slow- moving water?

Triss [41]

It have more energy picking up more particles in less time

5 0

3 years ago

Someone please help! Thxx

tangare [24]

Answer:

E, needs more info to be determined

Step-by-step explanation:

We know that Kai takes 30 minutes round-trip to get to his school.

One way is uphill and the other is downhill.

He travels twice as fast downhill than uphill.

This means that uphill accounts for 20 minutes of the round-trip and downhill accounts for 10 minutes of his trip.

However, even with this information, we do not know how far his school is.

In order to figure out how far away his school is, we would need more information about the speed at which Kai is traveling.

Simply knowing that he travels twice as fast downhill is not enough.

This question could only be solved by knowing how many miles Kai travels uphill or downhill in a given time.

3 0

3 years ago

Select interior, exterior, or on the circle (x - 5) 2 + (y + 3) 2 = 25 for the following point. (2, 3)

lozanna [386]

Standard equation of a circle: (x-h)² + (y-k)² = r² where (h, k) is the center and r is the radius. In the case of our equation here, (x-5)² + (y+3)² = 25, we can conclude that our circle has a center at (5, -3) and a radius of 5 units.

We can use the distance formula with the center (5, -3) and our point (2, 3) to see how far away they are...if the distance between them is less than the radius of the circle, it is on the interior. If it's equal, it's on the circle. If it's greater, it's on the exterior.

Distance = $\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Distance = $\sqrt{(-3-3)^2+(5-2)^2}$

Distance = $\sqrt{(-6)^2+3^2}$

Distance = $\sqrt{36+9}$

Distance = $\sqrt{45}\approx6.7082$

$6.7082>5,\ so\ (2,3)\ is\ on\ the\ \boxed{exterior}$

8 0

3 years ago

Subtract: (11w^3 + 15) – (-11w^3)​

Subtract: (11w^3 + 15) – (-11w^3)