Anthrax

With these problems, I like to change everything to 3D coordinates to make it easier for me to understand.

Currently, the room is a rectangular prism with important vertices of \((0,0,0)\) and \((12,10,8)\).

Now we should find out the gecko's current position. It shouldn't matter what side wall is chosen for the lil guy to be on, because the answer will be the same in this case, so we can choose the side wall closest to the origin, with an x-value of 0. Also depending on what you choose to be the front and back wall, this will affect the coordinates. I have illustrated how I chose such. Definitely not to scale.

In this sense, the Gecko's coordinates are \((0,0+1,8-1) = (0,1,7)\) and the fly's are \((12,10-1,0+1) = (12,9,1)\). With the tricky bit out of the way, we can now just use the 3D distance formula.

\(D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\), or in our case \(D = \sqrt{(x_F-x_G)^2+(y_F-y_G)^2+(z_F-z_G)^2}\)

\(D = \sqrt{(12-0)^2+(9-1)^2+(1-7)^2} = \sqrt{144+64+36} = \sqrt{244} = 2\sqrt{61}\)

(a) I want the midpoint value of the interval for the point estimate, so \(\frac{0.31+0.52}{2} = 0.415\)

(b) The poll's margin of error can be described as how far either end of the interval is from the point estimate, so \(|0.52-0.415| = 0.105\)

(c) estimate \(= \hat{p} * n = 0.415(140371) = 58253.965\)

A. \(\hat{p} = \frac{56}{80} = 0.7\)

B. estimated \(= \hat{p}*n = 0.7(138246) = 96772.2\)

Well, since this is inverse sine of a sine it cancels out to return to the given angle of \(\frac{3\pi}{8}\).

But, we should always check where stuff is defined with inverse trig functions and make sure \(\frac{3\pi}{8}\) is within it. The range of \(sin^{-1}\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\), and \(\frac{3\pi}{8}<\frac{\pi}{2}\), so the answer is \(\frac{3\pi}{8}\). 

A general Fibonacci sequence is defined as: \(a_n = a_{n-1} + a_{n-2}\)

We can use what was given, that \(a_5 = 15 \) and \(a_8 = 79\) to solve for \(a_1\)

\(a_6 + a_7 = a_8 \rightarrow a_6 + a_7 = 79\)

\(a_5 + a_6 = a_7 \rightarrow 19 + a_6 = a_7\)

Substituting for \(a_7 \) in the top equation , we can see \(a_6 + (a_6 + 19) = 79\)

\(2a_6 = 60 \rightarrow a_6 = 30\)

This part gets a lil repetitive

\(a_{4} + {19} = {30} \rightarrow a_{4} = {11}\)

\(a_{3} + {11} = {19} \rightarrow a_{3} = {8}\)

\(a_{2} + {8} = {11} \rightarrow a_{2} = {3}\)

\(a_{1} + {3} = {8} \rightarrow a_{1} = {5}\)

Therefore, the first entry of this sequence is 5.

Just to reformat your question for my aesthetic quick;

Simplify \((x+1)(2x+3)(3x+4) = (6x^{2}+5)(x-3)\) to the form \(0 = Ax^{2}+Bx+C\) where A, B, and C are positive integers with a greatest common divisor of 1. What is \(A+B+C\)?

\((x+1)(2x+3)(3x+4) = (6x^{2}+5)(x-3)\)

\((2x^2+3x+2x+3)(3x+4) = 6x^3-18x^2+5x-15\)

\((2x^2+5x+3)(3x+4)=6x^3-18x^2+5x-15\)

\(6x^3+8x^2+15x^2+20x+9x+12=6x^3-18x^2+5x-15\)

\(6x^3+23x^2+29x+12=6x^3-18x^2+5x-15\)

\(41x^2+24x+27=0\)

\(Ax^2 + Bx + C = 0\)

\(A = 41, B = 24, C = 27\)

\(A + B + C = 41+24+27=92\)

To start this problem, we should get a general idea of the probabilities of each product of the outcomes.

Let's make a table of the products.

The probability of each product then becomes:

\(P(1) = \frac{1}{16}, P(3) = \frac{2}{16}, P(8) = \frac{4}{16}, P(9) = \frac{1}{16}, P(24) = \frac{4}{16}, P(64) = \frac{4}{16}\)

For the decision of side dishes to be fair, we would want each dish to have a \(\frac{1}{4} \) or \(\frac{4}{16}\) probability of being prepared.

Since three products (8, 24, and 64) already have probabilities of \(\frac{4}{16}\), we would want those probabilities being assigned to only one dish each, and the other three products (1, 3, and 9) add up to \(\frac{4}{16}\) for the last dish. 

To rephrase, product of 8 goes to one dish, product of 24 goes to another, product of 64 goes to another, and products 1, 3, and 9 go to the last dish.

Answer B satisfies this.

I'm unsure if I'm interpreting this correctly but here goes lmao.

\(\frac{71}{5} * (6^7 - (-986533) * 3268358218) * 6*555535352 + (55555*636218*368*96492) * 5\)

\(\frac{71}{5} * (279936 + 3224343237878194) * 3333212112 + 6275354076445027200\)

\(\frac{71}{5} * 3224343238158130 * 3333212112 + 6275354076445027200\)

\(152613363072370508719241952 + 6275354076445027200\)

\(152613369347724585164269152\)

Another way of writing \(\lfloor x\rfloor = 4\) defines x values in the interval of \(4 \leq x < 5\)

If we multiply this interval by 4, we have that \(16\leq 4x < 20\).

In order to get the sum, we have to acknowledge that our 4x is still being floored, given \(\lfloor 4x \rfloor\). Therefore, only integer values within the interval can be a result of this. For example, \(x = 4.625\) satisfies that \(\lfloor x \rfloor = 4\). Then, \(\lfloor 4x \rfloor = \lfloor 4*4.625\rfloor = \lfloor 18.5 \rfloor = 18\)

The integers within \([16, 20)\) are 16, 17, 18, and 19. \(16+17+18+19= 70\).

If \(\lfloor x \rfloor = 4, \) the sum of all possible values of \(\lfloor 4x \rfloor\) is \(70\).

That's my bad, I was being cocky in what I currently had and not open-minded to his answer which I now realize to be true.