Determine whether the given set S is a subspace of the Vector space V.

(there are multiple)

2022-03-31

Determine whether the given set S is a subspace of the Vector space V.

(there are multiple)

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asked 2022-04-01

A penny of mass 3.1 g rests on a small 29.1 g block supported by a spinning disk of radius 8.3 cm. The coefficients of friction between block and disk are 0.742 (static) and 0.64 (kinetic) while those for the penny and block are 0.617 (static) and 0.45 (kinetic).

asked 2022-03-29

Sketch grapf of f(y)=y for dy\dt = ay+by^2,a>0,b>0,y>0

asked 2022-04-03

Use the method of your choice to factor the trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication.

6y^2 − 23y + 15

asked 2022-04-23

Given an sample of data set X which is normal-distributed to $N(\mu =240,\sigma )$, I want to find the $95\mathrm{\%}$ confidence interval of $max(\mu -200,0)$. As the $max(\mu -200,0)$ cut the normal-distribution curve at the point of 200, the sample of $max(\mu -200,0)$ is not more normal distributed. Therefore what is a reasonable $95\mathrm{\%}$ confidence interval?

asked 2022-03-28

- Suppose a fast-food restaurant can expect two customers every 3 minutes, on average. What is the probability that four or fewer customers will enter the restaurant in a 9-minute period?

asked 2022-05-12

$y\le 4$

asked 2022-04-25

How to solve a cyclic quintic in radicals?

Galois theory tells us that

$\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:

Let the roots be $\zeta}^{1},{\zeta}^{2},\dots ,{\zeta}^{10$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].

$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta}^{2}{x}_{3}+{\zeta}^{3}{x}_{4}+{\zeta}^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta}^{2}{x}_{2}+{\zeta}^{4}{x}_{3}+\zeta {x}_{4}+{\zeta}^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta}^{3}{x}_{2}+\zeta {x}_{3}+{\zeta}^{4}{x}_{4}+{\zeta}^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta}^{4}{x}_{2}+{\zeta}^{3}{x}_{3}+{\zeta}^{2}{x}_{4}+\zeta {x}_{5}\end{array}$

Once one has $A}_{0},\dots ,{A}_{4$ one easily gets $x}_{1},\dots ,{x}_{5$. It's easy to find $A}_{0$. The point is that $\tau$ takes $A}_{j$ to $\zeta}^{-j}{A}_{j$ and so takes $A}_{j}^{5$ to $A}_{j}^{5$. Thus $A}_{j}^{5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A}_{1}^{5$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have $A}_{1$ as a fifth root of a certain explicit complex number. Then one can express the other $A}_{j$ in terms of $A}_{1$. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.