Recent content by ferry2

Thanks a lot for the tips :).

So under your guidance the row reduced eshelon form of the matrix: A=\left( \begin{array}{cccc}1 &-1 & 6 & 0\\ 3 &-2 & 1 & 4\\ 1 &-2 & 1 &-2\\ 10 & 1 & 7 & 3\\ \end{array} \right) is \left( \begin{array}{cccc}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{array} \right)...

I don't wan't a solution I wan't only instructions how to solve this problem: Find a basis for the span: \vec{a_{1}}=(1,\,-1,\,6,\,0),\,\vec{a_{2}}=(3,\,-2,\,1,\,4),\,\vec{a_{3}}=(1,\,-2,\,1,\,-2),\,\vec{a_{4}}=(10,\,1,\,7,\,3)

http://math.fullerton.edu/mathews/n2003/hornermod.html is the Horner's method.

Thanks a lot. I solved by squaring and then I apply Horner's method. The real root's are -1 and -21.

Well \vec{AB}(8, a+1), \vec{AM}(7, \frac{15+a}{2}) \Rightarrow \vec{AM}\vec{AB}=56+\frac{1}{2}(a+1)(15+a) |\vec{AM}||\vec{AB}|=\sqrt{64+(a+1)^2}\sqrt{16+\frac{(15+a)^2}{4}} Then \vec{AM}\vec{AB}=|\vec{AM}||\vec{AB}|\frac{\sqrt{2}}{2} \Leftrightarrow...

Let the median is AM. I started with calculating the coordinates of the point M which is M(\frac{5+3}{2}, \frac{a+13}{2})\Rightarrow M(4, \frac{a+13}{2}). Then I calculate the dot product of \vec{AM} \vec{AB}. To be an angle \angle (\vec{AM}, \vec{AB})=45 degrees should \vec{AM}...

Points A (-3, -1), B (5, a), C (3, 13) are vertices of a triangle. Find the values ​​of parameter a for which the angle between AB and the median, passing through A is equal to 45 degrees.

Thanks for the replies. Already handled with these problems :cool:.

Can you tell me is my solution true of the next problem. Find center w_0 and radius R of the circle k, in which the transformation w=\frac{z+2}{z-2} converts the line l:\text{Im} z+\text{Re} z=0. Solution: 2 \to\infty -2i=(2)^*\to w_0...

Thanks you all :)

Looking for book "Physics and Applications of the Josephson Effect" Hello. I'm at the beginning of a project related to the Josephson effect and need the book "Physics and Applications of the Josephson Effect". I searched her long time, but I fail to find it. If anyone has it I would greatly...

It is possible there have been a typo. Thank you both.

Thanks a lot Ross Tang! Can you tell something about first equation?

Hi, friends :smile:. These are two equations, which were unable to resolve. Hope to help me. Note: this is not home, I just want to see how to resolve the equations. Thank answered. (2y-x+1)dx-(x-3y^2)dy=0 Find the common solution of the Euler's eqution: (2x+1)^2y''-2(2x+1)y'+4y=0...