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3 years ago

What is the slope of the line? y - 4= -7(2 – 6)

Mathematics

1 answer:

Serga [27]3 years ago

8 0

Answer:

-7

General Formulas and Concepts:

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate

y₁ - y coordinate

m - slope

Step-by-step explanation:

Step 1: Define function

y - 4 = -7(x - 6)

Step 2: Break function

Point (4, 6)

Slope m = -7

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1/3(m-1)=-5 Please help

dezoksy [38]

Answer:

m=-14 I hope this helps u

Step-by-step explanation:

6 0

3 years ago

Please help explanation if possible

Eva8 [605]

Answer:

612 adults

361 students

Step-by-step explanation:

To solve this question, set two equations:

Let x be number of adults and y be number of students.

As there are in total 937 people, the equation would be the sum of both adults and children:

$x+y=937$

$x=937-y$ ...... ( 1 )

As the total sale amount is $1109, the equation would be to add up the ticket fee:

$2x+0.75y=1,109$ ...... ( 2 )

Put ( 1 ) into ( 2 ):

$2(937-y)+0.75y=1,109$

$1874-2y+0.75y=1,109$

$-1.25y=-765$

$y=763/1.25$

$y=612$

Put y into ( 1 ):

$x=973-612$

$x=361$

Therefore there are 612 adults and 361 students.

5 0

3 years ago

Perimeter is the distance around a figure or region. a. true b. false

makkiz [27]

A. True

The sum of the distances round the shape is the Perimeter.

3 0

3 years ago

Read 2 more answers

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 cap $z=\sqrt{3-x^2-y^2}$ , 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:

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

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Draw a line representing the "run" and a line representing the "rise" of the line. State

Sonbull [250]

Answer:

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3 years ago