How many solutions are there to the system of equations graphed below if the lines are parallel

Mathematics

2 answers:

melamori03 [73]3 years ago

8 0

Answer:

Zero

C is the correct option.

Step-by-step explanation:

In the given graph, the given lines are parallel to each other and hence never intersect.

Now, we know that the intersection point (s) of two curves gives us the solution.

Here the lines are parallel to each other and hence do not intersect each other.

Thus, there would not be any intersection point.

Hence, the system of equations has no solution.

C is the correct option.

mash [69]3 years ago

3 0

If a consistent system has an infinite number of solutions, it is dependent . When you graph the equations, both equations represent the same line<span>. If a system has no solution, it is said to be </span>inconsistent<span> . The graphs of the </span>lines<span> do not intersect, so the graphs </span>are parallel<span> and there is no solution.

Sorry, this is a general answer. I could not view the graph below...</span>

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Evaluate the integral below, where C is the curver(t) = ‹sin(t),cos(t), sin(2t)›, 0 ≤ t ≤2π. (Hint: Observe that C lies on the s

nordsb [41]

We can compute the integral directly: we have

$\begin{cases}x(t)=\sin t\\y(t)=\cos t\\z(t)=\sin(2t)\end{cases}\implies\begin{cases}\mathrm dx=\cos t\,\mathrm dt\\\mathrm dy=-\sin t\,\mathrm dt\\\mathrm dz=2\cos(2t)\,\mathrm dt\end{cases}$

Then the integral is

$\displaystyle\int_C(y+9\sin x)\,\mathrm dx+(z^2+4\cos y)\,\mathrm dy+x^3\,\mathrm dz$

$=\displaystyle\int_0^{2\pi}\bigg((\cos t+9\sin(\sin t))\cos t-(\sin^2(2t)+4\cos(\cos t))\sin t+\sin^3t\bigg)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\cos^2t+9\cos t\sin(\sin t)-\sin^2(2t)\sin t-4\sin t\cos(\cos t)+\sin^3t\,\mathrm dt$

You could also take advantage of Stokes' theorem, which says the line integral of a vector field $\vec F$ along a closed curve $C$ is equal to the surface integral of the curl of $\vec F$ over any surface $S$ that has $C$ as its boundary.

In this case, the underlying field is

$\vec F(x,y,z)=\langle y+9\sin x,z^2+4\cos y,x^3\rangle$

which has curl

$\mathrm{curl}\vec F(x,y,z)=-\langle2z,3x^2,1\rangle$

We can parameterize $S$ by

$\vec s(u,v)=\langle u\cos v,u\sin v,2u^2\cos v\sin v\rangle=\langle u\cos v,u\sin v,u^2\sin(2v)\rangle$

with $0\le u\le1$ and $0\le v\le2\pi$ .

Note that when viewed from above, $C$ has negative orientation (a particle traveling on this path moves in a clockwise direction). Take the normal vector to $S$ to be pointing downward, given by

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle2u^2\sin v,2u^2\cos v,-u\rangle$