Anthrax

So, our current evidence that supports a difference exists between the first group and the second group is the difference in means being 3. So, looking at a line plot of random simulations, we want 3 to be an unlikely result, thereby showing that the difference in means did not happen by random chance. However, we do not see that in this case. 7/10 of the random simulations resulted in a difference between means in the two groups of or greater than three, so therefore the difference in the means of the two groups is not significant.

Incorrecto; this sentence contains no command form of the verb 'comer', instead it just reads 'you eat a lot' and I feel personally attacked tbh.

a) I will be substituting for x and simplifying.

\(\frac{4x^3+2x^2+6x+7}{2x+1} = 2x^2 + 3 + \frac{4}{2x+1}\)

\(\frac{4(1000)+2(100)+6(10)+7}{2(10)+1} = 2(100) + 3 + \frac{4}{2(10)+1}\)

\(\frac{4267}{21} = \frac{4267}{21}\)

(b) When plugging in \(x=-\frac{1}{2}\) to the denominators \(2x+1\) on both sides, we will end up dividing by zero, an undefined value.

(c) There's really two ways to do this. One is multiplying both sides by \(2x+1\) and showing the results. The other is dividing \(4x^3+2x^2+6x+7\) by \(2x+1\) through long division to show that it equals \(2x^2+3+\frac{3}{2x+1}\). However, I happen to hate long division with polynomials. So Ima do it tha otha way.

\(\frac{4x^3+2x^2+6x+7}{2x+1} = 2x^2 + 3 + \frac{4}{2x+1}\)

\(4x^3+2x^2+6x+7 = (2x+1) (2x^2 + 3 + \frac{4}{2x+1})\)

\(4x^3+2x^2+6x+7 = 4x^3+2x^2+6x+7\)

The right side multiplication looks intimidating, but it actually multiplies out that easily in one step.

Oh, and, see (b) for why it doesn't work for \(x = -\frac{1}{2}\)

When I started drawing this problem out, I was really focusing on both the aspects of "smallest possible sum" and "greatest common divisor" which for some reason combined in my head to think of using the least common multiple of the 3 sides that make the vertex be the number at the vertex. I think this would also create the smallest possible sum. I'm not entirely sure, though, lmao.

Here is my solution.

Please excuse my poor handwriting, bad lighting, and dirty whiteboard hehehe

Also this is where I do all my math if anyone is curious!!

To clarify, the drawing has visible aspects of the die in purple and hidden aspects of the die in orange. So each vertex becomes the least common multiple of each side that it touches. For example, \(a = lcm(1,3,5) = 15\).

So, my answer is 153. I'm not sure if this is THE lowest answer, but this is what I came up with.

I agree. The question states "She travels a certain distance on the first day, half that distance on the second day, a third that distance on the third day..."

The usage of the word 'certain' combined with a repeated 'that,' as in referring to the same certain distance over and over, implies the same setup to the problem for me.

Visualizing this problem can go a few different ways, but I'm just gonna keep it in simple geometry.

Like, really simple geometry lmao.

To explain the numbers, the horizontal 130 N is given to us. The downward force, or weight of the acrobat, is \(115 * 9.8 = 1127\).

Now, what is the magnitude of the force exerted by the floor on his hand? Well, it definitely isn't a normal force with two perpendicular forces. Oo! Then we can just use the resultant of the two forces. In other words, the magnitude of the force exerted by the floor on his hand is the hypotenuse of the above triangle in the drawing.

\(\sqrt{1127^2+130^2} = 1134.473N\)

\(F_n = mg\), so let's find the mass

We know the force, \(F\), and the acceleration, \(a\), so we can use \(F = ma\)

\(F = ma \rightarrow 1050 = m(0.785) \rightarrow m = 1337.58 kg\)

Now that we know the mass, we can go back to \(F_n\)

\(F_n = mg \rightarrow F_n = (1337.58)(9.8) \rightarrow F_n = 13108.3 N\)

I'm a lil confused. Are there no bounds to confine the area to? Otherwise, due to the nature of this function, the area under the graph but above the x-axis is \(\infty\).

Ima reformat da question right quick nawmsayn:

Find the positive value of \(n\) such that the equation \(9x^2+nx+1=0\) has exactly one solution in \(x\).

When working with quadratic equations and the question mentions something about a number of solutions, we definitely want to use the discriminant \(b^2-4ac\).

Let's review what the output of the discriminant tells you about the solutions to a quadratic equation.

If \(b^2-4ac > 0\), there are two solutions.

If \(b^2-4ac = 0\), there is one solution.

If \(b^2-4ac < 0\), there are no solutions.

Since we want only one solution, we want \(b^2-4ac = 0\). Where currently \(a = 9, b = n, c = 1\)

\(b^2 - 4ac = 0 \rightarrow n^2 - 4(9)(1) = 0\rightarrow n^2 = 36 \rightarrow n = \pm 6\)

The question asks for the positive value of \(n\), therefore \(n = 6\)