True or false !!! T or F

Can anyone help me with these problems I would really appreciate it

aksik [14]

Answer:

Starting from top left to right each row

1. 9

2. 17.5

3. 10.5

4. 22.5

5. 10

6. 16

7. 9

8. 19

9. 15

I hope that helps.

Step-by-step explanation:

5 0

2 years ago

Determine the center and radius of the following circle equation: x2 + y2 – 10x + 8y + 40 = 0

antiseptic1488 [7]

Answer:

The center of this circle is (5, -4) and the radius is 1.

Step-by-step explanation:

First regroup these terms according to x and y :

x^2 - 10x + y^2 + 8y = -40

Next, complete the square for x^2 - 10x: x^2 - 10x + 5^2 - 5^2.

and the same for y^2 + 8y: y^2 + 8y + 16 - 16

Substituting these results into x^2 - 10x + y^2 + 8y = -40, we get:

x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40.

Next, rewrite x^2 - 10x + 25 and y^2 + 8y + 16 as squares of binomials:

Then x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40 becomes:

(x - 5)^2 + (y + 4)^2 - 25 - 16 = -40, or:

(x - 5)^2 + (y + 4)^2 = 1

This equation has the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Matching like terms, we get h = 5, k = -4 and r = 1.

The center of this circle is (5, -4) and the radius is 1.

6 0

3 years ago

Read 2 more answers

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

$\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV$

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

$\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}$

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$