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trasher [3.6K]
4 years ago
12

What is the constant of proportionality in the equation y = 2.5x?

Mathematics
1 answer:
Oksi-84 [34.3K]4 years ago
3 0

Answer:

k = 2.5

Step-by-step explanation:

The standard form of a proportional equation is

y = kx ← k is the constant of proportion

y = 2.5x ← is in this form

with k = 2.5

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What is the approximation for 78.8×6.8
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78.8x6.8 would be 535.84
7 0
2 years ago
Antonia and Carla have the same percentage of green marbles in their bags of marbles. Antonia has 4 green green marbles and 16 t
natulia [17]
Given that Antonia and Carla have the same percentage of marbles and Antonia has 4 green marbles out of 16 marbles , to calculate the number of marbles that Carla has that aren't green we proceed as follows;
let the total number of Carla's marbles be x;
the percentage of green marbles that Antonia has is:
4/16×100=25%

since they have the same percentage then:
number of green marbles Carla has is 10:
this is 25% of x, hence:
10/x×100=25
1000/x=25
1000=25x
thus
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7 0
4 years ago
Graph the following<br> y−2=2/3(x+4)
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4 0
3 years ago
Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −
Korolek [52]

The vector field

\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k

has curl

\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath

Parameterize S by

\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k

where

\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}

with 0\le u\le3 and 0\le v\le2\pi.

Take the normal vector to S to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k

Then by Stokes' theorem we have

\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv

which has a value of 0, since each component integral is 0:

\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0

4 0
3 years ago
Help please :-))))))))
ira [324]
The answer is A....a=5 b=30
4 0
3 years ago
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