Three fair coins are tossed, and we let $X_1$, denote the number of heads that appear. Those coins that were heads on the first trial (there were $X_1$, of them) we pick up and toss again, and now we let $X_2$, be the total number of tails, including those left from the first toss. We toss again ...

Should one approach by coordinates or by euclidean geometry? By pure geometry, I am not able to solve.

I was computing the homology of $S^3-\coprod_{i=1}^4 I_i$, where $I_i=[0,1]$ for all $i$ (they are being identified with an embedding). Intuitively, this should be homotopy equivalent to $S^1$, since removing one interval gives something homotopic to $\mathbb{R}^3$, removing another one gives an ...

How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$? My attempt : I got $2$, that is $ z=-2, z= +2$ , $|z| = {\sqrt{ 2^2+1}}$, $|z| = {\sqrt{(-2^2) +1}}$ Is it true ?

I need help with determining the basis of $U_1 \cap U_2$ in the following problem: Let $V=\mathbb{R}^4$. ${U_1} = \left\{ {\left( {\begin{array}{*{20}{c}} {{x_1}} \\ {{x_2}} \\ {{x_3}} \\ {{x_4}} \end{array}} \right)\left| {{x_1} - {x_2} + {x_3} - 3{x_4} = 0} \right.} \right\}$ and $...

$17!$ is equal to $$35568x428096y00$$ Both x and y, are digits. Find x,y. So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$ If there`s a product of $(2*5)^3$ Then this number has $3$ zeros at the end, so $y=0$ How do I find the $x$ now?