Mathematics, 06.06.2020 21:59 maddoxlachowski

Suppose two dice are rolled. Consider the following events. A: the red die shows 4; B: the numbers add to 6; C: at least one of the numbers is 1; and D: the numbers do not add to 11. Either the numbers add to 6, or they add to 11, or at least one of them is 1.

How many elements does it contain?

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Answer from: Quest

honestly bro aint no one got time to answer these dumb question just do it p***y

step-by-step explanation:

Answer from: Quest

0lz try yourself sorry for not answering

Answer from: Quest

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.

$\bf \textit{hyperbolas, vertical traverse axis } \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} [-0.35em] \rule{34em}{0.25pt} \begin{cases} h=0\\ k=0\\ a=\sqrt{5}\\ c=3 \end{cases}\implies c=\sqrt{a^2+b^2}\implies 3=\sqrt{(\sqrt{5})^2+b^2}\implies 3^2=5+b^2$

$\bf 3^2-5=b^2\implies 9-5=b^2\implies 4=b^2\implies \sqrt{4}=b\implies \boxed{2=b} [-0.35em] ~\dotfill \cfrac{(y-0)^2}{(\sqrt{5})^2}-\cfrac{(x-0)^2}{2^2}=1\implies \cfrac{y^2}{5}-\cfrac{x^2}{4}=1$

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Suppose two dice are rolled. Consider the following events. A: the red die shows 4; B: the numbers a...