PartialMathematician

Nota bene: positive means greater than 0, and negative means less than 0. Positive and negative real numbers do not include 0. If you wanted non-positive and non-negative, you need to restate your question. Otherwise, here is the solution. 

A) To have f(t) be positive, either \((t-4)\) and \((t+9)\) must both be positive or \((t-4)\) and \((t+9)\) must both be negative. In order to have (t-4)(t+9) be positive, since (t-4) < (t+9), we need (t-4) to be greater than 0. We have the inequality \(t-4>0\), which simplifies into \(t > 4\). We cannot have t = 4 because (4-4) = 0, and anything multiplied to 0 is 0 (which is not positive). We can also plug in some values of \(t\) and try them out. If t = 5, we have \((5-4)\cdot(5+9)\), which simplifies into \((1)\cdot(14)\), which is greater than 0. 

In order to have (t-4) and (t+9) to both be negative, since (t+9) > (t-4), we need (t+9) to be less than 0. We have the inequality \(t+9<0\), which simplifies into \(t<-9\). We cannot have t = -9 because (-9+9) = 0, and anything multiplied by 0 is 0. If we try t = -10, we have \((-10-4)\cdot(-10+9)\), which simplifies into \((-14)\cdot(-1)\). Since the minus signs cancel out each other, we are left with 14, which is greater than 0. 

Combining both inequalities \(t>4\) and \(t<-9\), we have the answer: \(\boxed{t<-9}\) or \(\boxed{t>4}\) (or \(\boxed{t \in (-\infty, -9)\cup(4, \infty)}\)in intervals). 

B) To have f(t) be negative, either \((t-4)\) is positive and \((t+9)\) is negative or \((t-4)\) is negative and \((t+9)\) is positive. If (t-4) is positive, then (t+9) must also be positive because (t+9) > (t-4), so we can eliminate the first scenario. Since (t+9) > (t-4), to have (t+9) be positive, \(t\) must be more than -9. We cannot have t = -9 because (-9+9) = 0, and anything multiplied to 0 is 0. \(t\) can also not be more than 4 because (t-4) would then equal 0 or be positive, which is not what we want. Combining the inequalities \(t>-9\) and \(t<4\), we have the answer: \(\boxed{-9 .

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- PartialMathematician

Could you please specify which sides the six points A,B,C,D,E, and F are on? For example, A and B are on side XY. If by chance there is a diagram, could you also attach it to your post? 

Thanks, and with that piece of information, I should be able to answer your question. 

...Is this some trick question?

 Statement 8 is incorrect...It claims 2 = 1...

Besides, qeustions not related to math should be considered "off-topic"...

I am not sure I understand your question...

Maybe gray does not appeal to you...

Let's C, it's D

If you did not know already, "μ" is the greek letter "mu", and it stands for micro, or \(10^-6\) (10^-6).

The "η" is the greek letter "eta", and it stands for the greek letter "eta", and it stands for the "Dirichlet eta function".  In mathematics, in the area of analytic number theory, the Dirichlet eta function is defined by the following Dirichlet series, which converges for any complex number having real part > 0: 

\({\displaystyle \eta (s)=\sum _{n=1}^{\infty }{(-1)^{n-1} \over n^{s}}={\frac {1}{1^{s}}}-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots }\)

...Except this is not analytic number theory, and "μ" simply represents tthe mean of X and "η" simply represents the median of X. Questions b) and c) tell us that "μ"  = 1 and "η" = \(\dfrac{13}{12}\). (The top part was a joke...)

\(3\cdot μ = 3\) and \(3 - \dfrac{13}{12} = 1\dfrac{11}{12}\). Question d) asks us to find the probabliltiy that \(X\lt 1\dfrac{11}{12}\). \(X\) can only be greater than \(1\dfrac{11}{12}\) when \(f(x) = \dfrac{3-x}{4}\). I believe you can go on from here and find the probabililty that \(X\lt 1\dfrac{11}{12}\). If you need more help, you could re-state the question or ask for more information in forums. 

- PartialMathematician

I used to think that the percent sign (%) was made of a "1" and two ''0"s, which can be formed into 100. So a percent (%) of something is something out of 100. 

21% = \(\dfrac{21}{100}.\) 

In LaTex, it would be​ \(y = 1, 200\cdot 1.02^x\)

If you want to use "-->" or "==>" in LaTeX, it is $\Rightarrow$.