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Alinara [238K]
3 years ago
11

Graph g(x), where f(x) = 2x − 5 and g(x) = f(x + 1).

Mathematics
2 answers:
vivado [14]3 years ago
7 0

Answer:

It's a line labeled  that passes through points  and ,

Step-by-step explanation:

To graph , first we have to find that function by replacing .

The  is defined as the first function but with its variable increased by 1 unit, instead of , we write  as its variable.

So, if we graph this function we'll have something like the image attached. It's a line labeled  that passes through points  and , which is showed on the graph.

Step-by-step explanation:

pav-90 [236]3 years ago
5 0

Answer:

B.) a line labeled g(x) that passes through points 0, negative 3 and 4, 5

Step-by-step explanation:

I graphed both equation on the graph below to find the description of the graph of g(x).

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Step-by-step explanation:

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The ordered pairs (t, 36), (5, 45) represent a proportional relationship. Find the value of t.
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The answer for t is 4 and the proportional relationship is 9
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Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of R by

\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
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