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scZoUnD [109]
3 years ago
5

19/5 to mixed number

Mathematics
2 answers:
Dovator [93]3 years ago
8 0

Answer:

3 4/5

Step-by-step explanation:

5 x 3= 15

19 - 15= 4

19/5 = 3 4/5

Liono4ka [1.6K]3 years ago
5 0

Answer:

3 4/5

Step-by-step explanation:

19/5 there are 3 wholes of 5.

5,10,15

3 wholes later we are left with 4

You just write it as 4/5. Meaning, the 4 is just left.

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Solve the problem, calculate the line integral of f along h
Over [174]
The curve \mathcal H is parameterized by

\begin{cases}X(t)=R\cos t\\Y(t)=R\sin t\\Z(t)=Pt\end{cases}

so in the line integral, we have

\displaystyle\int_{\mathcal H}f(x,y,z)\,\mathrm ds=\int_{t=0}^{t=2\pi}f(X(t),Y(t),Z(t))\sqrt{\left(\frac{\mathrm dX}{\mathrm dt}\right)^2+\left(\frac{\mathrm dY}{\mathrm dt}\right)^2+\left(\frac{\mathrm dZ}{\mathrm dt}\right)^2}\,\mathrm dt
=\displaystyle\int_0^{2\pi}Y(t)^2\sqrt{(-R\sin t)^2+(R\cos t)^2+P^2}\,\mathrm dt
=\displaystyle\int_0^{2\pi}R^2\sin^2t\sqrt{R^2+P^2}\,\mathrm dt
=\displaystyle\frac{R^2\sqrt{R^2+P^2}}2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt
=\pi R^2\sqrt{R^2+P^2}

You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function f:\mathbb R^n\to\mathbb R, we have gradient \nabla f:\mathbb R^n\to\mathbb R^n. The theorem itself then says that the line integral of \nabla f(x,y,z)=\mathbf f(x,y,z) along a curve C parameterized by \mathbf r(t), where a\le t\le b, is given by

\displaystyle\int_C\mathbf f(x,y,z)\,\mathrm d\mathbf r=f(\mathbf r(b))-f(\mathbf r(a))

Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.

But this isn't the case: we're integrating f(x,y,z)=y^2, a scalar function.
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