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3 years ago

Determine if the following system onf equations has no solutions, infinite solutions, or one solution

1 answer:

5 0

Let’s make them both slope intercept equations (y = mx + b)
-5x + 2y = 5
2y = 5 + 5x
y = 5/2x + 5

5x - 2y = -5
5x = 2y - 5
2y = 5 + 5x
y = 5/2x + 5

Same results, therefore the equations are the same.

Solution: infinite solutions

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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

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

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago

Find the vertex of the function h(x) = x2 – 4x – 21

lakkis [162]

Answer: (2, -25)

Step-by-step explanation:

The x coordinate of the vertex is $x=-\frac{-4}{2(1)}=2$ .

When x=2, $h(2)=2^2 - 4(2)-21=-25$ .

So, the vertex is (2, -25).

7 0

2 years ago

What is 6 divided by 25.2

mr_godi [17]

6/25.2
Multiply both sides by 5 so 25.2 is a whole number:
30/126
and then simplify by dividing by 6:
5/21 which is equal to 0.2381.

Hope this helps :)

6 0

3 years ago

J is the midpoint of hk. what is hj, jk, and hk

Lyrx [107]

Hj and jk are the same length line segments ( because the midpoint divides a line into two equal parts)
So hj = jk.

hk is the line segment which has the mid point j. It is the double of hj or jk. It can be the sum of hj and jk.
hj + jk = hk
or
2 * hj = hk
or
2 * jk = hk

4 0

3 years ago

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What is the expanded form of this?

ollegr [7]

Answer:

The first one

Step-by-step explanation:

...................

7 0

3 years ago

Read 2 more answers