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3 years ago

Determine the discriminant for the following function: x2+11x=12

Mathematics

1 answer:

Serga [27]3 years ago

6 0

Answer: 169

First make equal to zero

You substitude1=a,11=b,c= -12 in the equation x^2-4ac

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Find the area of the shaded region in each of the following.

zysi [14]

Answer:

Step-by-step explanation:

a) A = ½(5 - -1)(5) = 15 units²

b) A = ½(4 - 2)(4) = 4 units²

c) A = ½(6 - -3)(3) = 13.5 units²

d) A = ½(4 - -1)(8) = 20 units²

5 0

3 years ago

What is the domain of the following function: f ( x ) = √ x + 4/ x − 2

Tju [1.3M]

Answer:

(-∞,+∞) interval notation

it is not clear if the square root on the x only or the four too.

the answer above is if the equation is :

√x +4/x -2

5 0

3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$