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$\frac{20b}{{\left(4b^3\right)}^3}$

Geometric probability
Inside a square of side 2 units , five points are marked at random. What is the probability that there are at least two points such that the distance between them is at most $\sqrt{2}$ units?

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find the area of the region between Z = - 0.21 and Z 1.1 under the standard normal curve 

Negative Binomial Vs Geometric
So I am trying to get the difference between these two distributions. I think I understand them but the negative binomial has me a bit confused.
The Geometric is the probability of some amount of successes before the first failure.
The Negative binomial is the portability of some amount of successes before a specified number of failures? for example the number of successes before the 8th failure?

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Prove that
${\displaystyle 3{(\sin \theta -\cos \theta )}^4+6{(\sin \theta +\cos \theta )}^2+4({\sin }^6\theta +{\cos }^6\theta )-13=0}$

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${\displaystyle \frac{4x^2-x-23}{x-1}}$
Use polynomial long division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x)

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Use Laplace transform to solve the initial-value problem
${\displaystyle y=3,y\left(0\right)=1,y^{\prime }\left(0\right)=1}$

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Suppose $\stackrel{\to <!--\; \to \; -->}{v}$ and $\stackrel{\to <!--\; \to \; -->}{w}$ are two vectors parallel to the plane
$x+2y+3z=7.$
Suppose furthermore that $\stackrel{\to <!--\; \to \; -->}{v}$ is perpendicular to $\stackrel{\to <!--\; \to \; -->}{w}$ ,
$\Vert v\Vert =3,\Vert w\Vert =4.$
How would you go about answering this question? I always reach a dead end when I try to solve $\stackrel{\to <!--\; \to \; -->}{v}$ and $\stackrel{\to <!--\; \to \; -->}{w}$ that satisfy the given information, I just don't know how to go about it. I tried to draw the plane and two parallel vectors, then I know the length of the cross product would be 12. Then what do I do now? How could you solve this to find vectors $\stackrel{\to <!--\; \to \; -->}{v}$ and $\stackrel{\to <!--\; \to \; -->}{w}$?

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Determine the value of $k$ such that the image of circle $|z-<!--\; -\; -->1|=k$ by complex function $f(z)={\displaystyle \frac{z-<!--\; -\; -->3}{1-<!--\; -\; -->2z}}$ is a line.
I substitute $z$ in first equation by $f(z)$ then found that every $k$ makes the equation in form of $az+b$ which is a linear equation. Is this true? But I really doubt. Help me

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