y''+4y= f(t); y(0)= 1, y''(0)=0 , with

f(t) = { 0 for 0≤t<4 , 3 for t≥4}

2022-04-30

y''+4y= f(t); y(0)= 1, y''(0)=0 , with

f(t) = { 0 for 0≤t<4 , 3 for t≥4}

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asked 2020-12-30

A majorette in a parade is performing some acrobatic twirlingsof her baton. Assume that the baton is a uniform rod of mass 0.120 kg and length 80.0 cm.

With a skillful move, the majorette changes the rotation ofher baton so that now it is spinning about an axis passing throughits end at the same angular velocity 3.00 rad/s as before. What is the new angularmomentum of the rod?

asked 2022-06-30

Find the transformation matrix:

$F:{\mathbb{R}}_{3}[x]\text{}\mathbb{]}\to {\mathbb{R}}_{3}[x]$

$F(v)=\frac{{d}^{2}v}{d{v}^{2}}$

Basis: $1,x,{x}^{2},{x}^{3}$ and ${\mathbb{R}}_{3}[x]$ - the set of all third degree polynomials of variable $x$ over $\mathbb{R}$ Assume that all coefficients of the polynomials are $1$

The first thing that springs to my mind is to calculate this derivative by hand, and so we got

$\frac{{d}^{2}y}{d{y}^{2}}=2+6x$

Now, we need to put these values - $2$ and $6$ in such a matrix that - when multiplied by the basis vector -will give us $2+6x$ But there are many ways I can think of, for example

$\left[\begin{array}{cccc}0& 0& 2& 0\\ 0& 0& 0& 6\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Or maybe

$\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 6& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Because both of them, when multiplied by$\left[\begin{array}{c}1\\ x\\ {x}^{2}\\ {x}^{3}\end{array}\right]$

Will give the correct answer. Thus, what is the correct way to solve this?

$F:{\mathbb{R}}_{3}[x]\text{}\mathbb{]}\to {\mathbb{R}}_{3}[x]$

$F(v)=\frac{{d}^{2}v}{d{v}^{2}}$

Basis: $1,x,{x}^{2},{x}^{3}$ and ${\mathbb{R}}_{3}[x]$ - the set of all third degree polynomials of variable $x$ over $\mathbb{R}$ Assume that all coefficients of the polynomials are $1$

The first thing that springs to my mind is to calculate this derivative by hand, and so we got

$\frac{{d}^{2}y}{d{y}^{2}}=2+6x$

Now, we need to put these values - $2$ and $6$ in such a matrix that - when multiplied by the basis vector -will give us $2+6x$ But there are many ways I can think of, for example

$\left[\begin{array}{cccc}0& 0& 2& 0\\ 0& 0& 0& 6\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Or maybe

$\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 6& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Because both of them, when multiplied by$\left[\begin{array}{c}1\\ x\\ {x}^{2}\\ {x}^{3}\end{array}\right]$

Will give the correct answer. Thus, what is the correct way to solve this?

asked 2021-12-15

Find the first and second derivatives of the function.

$g\left(t\right)=2\mathrm{cos}t-3\mathrm{sin}$

asked 2022-05-01

Rotation matrix to construct canonical form of a conic

$C:9{x}^{2}+4xy+6{y}^{2}-10=0.$

I've found C is a non-degenerate ellipses (computing the cubic and the quadratic invariant), and then I've studied the characteristic polynomial

$p\left(t\right)=\mathrm{det}\left(\begin{array}{cc}9-t& 2\\ 2& 6-t\\ & \phantom{\rule{0ex}{0ex}}\end{array}\right)$

The eigenvalue are ${t}_{1}=5,{t}_{2}=10$, with associated eigenvectors $(-1,2),(2,1)$. Thus I construct the rotation matrix R by putting in columns the normalized eigenvectors (taking care that $det\left(R\right)=1$):

$R=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& 2\\ -2& 1\\ & \phantom{\rule{0ex}{0ex}}\end{array}\right)$

Then $(x,y)}^{t}=R{({x}^{\prime},{y}^{\prime})}^{t$, and after some computations I find the canonical form

$\frac{1}{2}{x}^{\prime 2}+\frac{4}{5}{y}^{\prime 2}=1.$

asked 2022-05-15

12 ≡ 15 mod 4

8 ≡ 21 mod 3

25 ≡ 60 mod 7

62 ≡ 30 mod 5

35 ≡ 95 mod 12

asked 2021-06-11

Complete the two-way table.

asked 2021-08-07

To calculate: The simplified form of the expression , $\sqrt{\frac{8.0\times {10}^{12}}{2.0\times {10}^{4}}}$