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3 years ago

8

the equation of a line is y = 3. write an equation in slope-intercept form of a line parallel to y = 3 that passes through (0, 6

). x = 3 x = 6 y = 3x y = 6

1 answer:

Pavel [41]3 years ago

6 0

Parallel lines have equal gradients;
Gradient of line 2 is thus 0
y=c
Replacing value of y=6 gives
c=6
Equation of the line is thus y=6

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What are the domain and range of the function?<br> f(x) = ^3 square x - 3

DerKrebs [107]

Step-by-step explanation:

$f(x) = \sqrt[3]{x - 3}$

The domain of a function are the values of $x$ that you can plug into the function $f(x)$ .

Values inside of a root must be non-negative, which means that $x - 3$ must be greater than or equal to zero. We can set up an equation to find the domain:

$x - 3 \geq 0$

$x \geq 3$

With this, we know the domain of the function is $[3, \inf)$ .

The range of a function are the values that $f(x)$ can have. Since the equation is a cube root, the value will always be non-negative, meaning the range of the function is $[0, \inf)$ .

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3 years ago

One way to show that two triangles are similar is to show that

avanturin [10]

Answer:

show that the three sets of corresponding sides are in proportion. If the three sets of corresponding sides of two triangles are in proportion, the triangles are similar.

Step-by-step explanation:

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3 years ago

Heather hits a tennis ball upward with an initial speed of 32 feet per second. After how many seconds does the ball hit the grou

Mrrafil [7]

set h=0 and solve for <span>t

</span>So: 0 = 32t - 16^2

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4 years ago

Read 2 more answers

Tom [10]

Answer: the function g(x) has the smallest minimum y-value.

Explanation:

1) The function f(x) = 3x² + 12x + 16 is a parabola.

The vertex of the parabola is the minimum or maximum on the parabola.

If the parabola open down then the vertex is a maximum, and if the parabola open upward the vertex is a minimum.

The sign of the coefficient of the quadratic term tells whether the parabola opens upward or downward.

When such coefficient is positive, the parabola opens upward (so it has a minimum); when the coefficient is negative the parabola opens downward (so it has a maximum).

Here the coefficient is positive (3), which tells that the vertex of the parabola is a miimum.

Then, finding the minimum value of the function is done by finding the vertex.

I will change the form of the function to the vertex form by completing squares:

Given: 3x² + 12x + 16

Group: (3x² + 12x) + 16
Common factor: 3 [x² + 4x ] + 16
Complete squares: 3[ ( x² + 4x + 4) - 4] + 16
Factor the trinomial: 3 [(x + 2)² - 4] + 16
Distributive property: 3 (x + 2)² - 12 + 16
Combine like terms: 3 (x + 2)² + 4

That is the vertex form: A(x - h)² + k, whch means that the vertex is (h,k) = (-2, 4).

Then the minimum value is 4 (when x = - 2).

2) The othe function is <span>g(x)= 2 *sin(x-pi)
</span>

The sine function goes from -1 to + 1, so the minimum value of sin(x - pi) is - 1.

When you multiply by 2, you just increased the amplitude of the function and obtain the new minimum value is 2 (-1) = - 2

Comparing the two minima, you have 4 vs - 2, and so the function g(x) has the smallest minimum y-value.

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4 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:

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

4 0

4 years ago