Anthrax

Integrating \(f'(x)\) with respect to \(x\) yields

\(f(x) = -\cos(x) + C\),

where \(C\) is a constant we can solve for given the initial condition. Because we are given \(f(0) = 10,\) we can say

\(10 = f(0) = -\cos(0) + C = -1 + C\).

This implies that \(C = 11.\) Hence,

\(f(x) = -\cos(x) + 11.\)

Let \(a_n\) denote the \(n\)th term of this arithmetic sequence. There are 27 terms occurring from \(a_5 = 9\) to \(a_{32} = 60\). Assuming a constant common difference between terms, then the terms increase by \(\frac{60-9}{27} = \frac{17}{9}\) with each \(n\) step. That is, \(a_{n+1} = a_n + \frac{17}{9}\).

There are 18 terms occurring from \(a_5 = 9\) to \(a_{23}.\) Then, there would be 18 additions of \(\frac{17}{9}\) from \(a_5\) to \(a_{23}\). That is,

\(a_{23} = a_5 + 18\left(\frac{17}{9}\right) = 9 + 34 = 43\).

Excellent use of results from Number Theory and Algebra, Max!!!

Observe that we can say

\(\displaystyle\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \cdots = \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{4}+\cdots\right) = \frac{1}{3}\sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n.\)

Given this, our common ratio \(r\) is \(r = \frac{1}{2}\). Then, by the well-known infinite geometric series formula, we have

\(\displaystyle \frac{1}{3} \sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n = \frac{1}{3} \cdot\frac{1}{1-\frac{1}{2}} = \frac{2}{3}.\)

I think that the key sentence in their answer is "Consider ordering the nine pairs by the time they are first complete." This is tricky to think about, but, in a combinatorial context, this does not skew the uniform randomness of the selection of cards. It's less that you gain the paired card with a draw of any card, but more so that the 7 N's and 2 J's serve as a model for the pairs that appear during Mario's selection. That is, the "position of the first J" is just analogous to the first point during Mario's drawing of cards whereby he has a pair of Jokers in front of him.

The general quadratic equation for a projectile's free fall is

\(p(t) = -at^2 + v_0t + h,\)

where \(a\) represents acceleration due to gravity, \(v_0\) is the initial (launch) velocity of the projectile, and \(h\) is our starting height.

In terms of feet, the acceleration of a projectile due to gravity (on Earth) comes out to \(32.17 \frac{\text{ft}}{\text{s}^2}\). I'll assume from your statement that there's no initial velocity present. Then, with a starting height of 6 feet, our quadratic equation will take the form

\(p(t) = -32.17t^2 +6.\)

Classic complete the square problem ^_^

If you expand the right side, we have

\(x^2 - 3x + 2 = x^2 - 2kx + k^2 + p\).

Because like terms have to match on both sides, we garner \(-3x = -2kx \implies k = \frac{3}{2}\).

Now, we need \(2 = k^2 + p.\) Substituting the value for \(k\) we just found, we get \(2 = \left(\frac{3}{2}\right)^2 + p \implies 2 = \frac{9}{4} \implies p = -1/4\).

So, we can conclude the answer \(d.) -1/4\).

I agree with your rationale for part a).

As for part b), you can apply the same logic of breaking each draw down. For the first two draws, you would have the same probability of \(\frac{2}{9}\) for drawing two black balls. The probability of having one or more white balls at this point is a complement of your desired event of having two black balls, so the probability of having one or more white balls is \(1 - \frac{2}{9} = \frac{7}{9}\). In other words, you will be redrawing \(\frac{7}{9}\) of the time. The probability of drawing two white balls would be the same as drawing two black balls because there are 5 of each, thus P(both white balls) = \(\frac{2}{9}\). Thus, the only other possibility of drawing exactly one white ball and one black ball would be \(1 - \frac{2}{9} - \frac{2}{9} = \frac{5}{9}\). 

Say you drew one white ball in your first drawing. This happens \(\frac{5}{9}\) of the time. If you put the white ball back, your probability of drawing a second black ball from the 9 with 4 black balls would be \(\frac{4}{9}\), so this probability is \(\frac{5}{9}*\frac{4}{9} = \frac{20}{81}\).

Say you drew two white balls in your first drawing. This happens \(\frac{2}{9}\) of the time. If you put both balls back, your probability of drawing two black balls would be the same as the initial drawing, since you have 5 black and 5 white in the bag now. Thus, this probability is \(\frac{2}{9}*\frac{2}{9}=\frac{4}{81}\).

Using the sum rule, as these are all disjoint events, and including the probability we chose two black balls on the first draw, we arrive at the probability being \(\frac{2}{9} + \frac{20}{81} + \frac{4}{81} = \frac{14}{27}\). This could be wrong I gotta get to class lmao

Since we know that their largest rug is a similar rectangle, i.e. having the same proportional shape to their smallest rug, there are two possibilities for the dimensions of the largest rug: either the largest rug's 18ft side is similar to the 2ft side or the 3ft side. This gives us an 18ft by \(x\) ft rectangle and a \(y\) ft by 18 ft rectangle. Using the original proportions, we can now solve for \(x \) and \(y \):

\(\frac{3}{2}=\frac{x}{18} \rightarrow x = 18* \frac{3}{2} \rightarrow x = 27\)

\(\frac{2}{3} = \frac{y}{18} \rightarrow y = 18 * \frac{2}{3} \rightarrow y = 12\)

Thus, the company's largest rug dimensions are either 18 ft by 27 ft or 12 ft by 18 ft. Hope this helps!

There's a couple of ways to do this problem, I'll do a rough-estimate solution and an algebraic one.

A simple solution to estimate how long it will be until the car's value is $12,000 is to simply depreciate the car's current value year-by-year by 18%, or 82% of its current value until the car is worth less than $12,000. In other words, we can keep multiplying the car's value by 0.82 until it is less than $12,000:

2011: $35000, 2012: $28700, 2013: $23534, 2014: $19297.9, 2015: $15824.3, 2016: $12975.9, 2017: $10640.2

Since its value in 2017 is now less than $12000, we know that it would take between 5-6 years for the depreciated car's value to be worth $12,000.

An exact solution will require an inkling of algebra. We can model what the car will cost, \(C\), at any point in \(t\) years given the initial price, \(P\), and decay rate, \(R\), using an exponential formula \(C = P * (1-R)^t\). Plugging in our givens, we have \(C = 35000(0.82)^t\). We want to know when the depreciated cost of the car is worth $12000, so we'll be solving for \(t\) with a fixed \(C\). Mathematically, we have \(12000 = 35000(0.82)^t\), which can be solved as follows:

\(12000 = 35000(0.82)^t\)

\(\frac{12}{35}=0.82^t\)

\(\log_{0.82} (\frac{12}{35})=t\)

\(t \approx 5.39399\)

So, we can see that it would be about 5.4 years from 2011 until the car's value is worth $12,000. This answer agrees with our rough estimate of between 5-6 years in our earlier approximation. Hope this helps :)

I agree with the other answer, this question is a tad confusing.

If we interpret it as each number being a number instead (weird sentence, I know), 100 would actually be the smallest three-digit number in Pascal's triangle. It is the second number in the 99th row (or 100th, depending on who you ask), or \(\binom{100}{1}\)

Thanks much!

Thank you Ginger!!!! <3

I guess I'll solve the second question then hahaha

To restate the important distinction in the question, \(g(x)\) is the table and \(f(x)\) is the graph.

Let's consider the validity of each option:

A: "\(f(x) \) and \(g(x)\) have the same y-intercept." For this to be true, we would want \(f(0) = g(0)\). The y-intercept of \(g(x)\) can be found in the table at \(g(0)\), which is equal to 2. Looking at the graph of \(f(x)\), we are unable to see the y-intercept at \(f(0)\), but it is definitely less than -6. Therefore, \(f(0) \neq g(0)\) and this is not true.

B: "\(f(x) \) and \(g(x)\) intersect at two points." For this to be true, we need two points from each graph that are equivalent to each other. In other words, at two \(x\) inputs, \(f(x)\) and \(g(x)\) give the same output. Let's go through each value from the table of \(g(x)\) and see if it matches \(f(x)\).\(g(-2) \neq f(-2), g(0) \neq f(0), g(2)=f(2), g(4) \neq f(2), \)and \(g(6) \neq f(6)\). We can see that \(f(x) \) and \(g(x)\) intersect at one point, but not at two. Therefore, this is not true.

C: "\(f(x)\) is greater than \(g(x)\) for all values of \(x\)." We can quickly find a counterexample for this. At \(x=4\), \(g(4) = 2\) and \(f(4) = 0\). Therefore, \(f(4) \ngtr g(4)\) and this is not true.

D: "\(f(x) \) and \(g(x)\) have a common x-intercept." We learned from looking at B that \(g(2) = f(2)\), wherein both equaled a y-value of \(0\). Therefore, this is true.

Our answer to this question is then the fourth option, or as I have labeled it, D.

\(F_s = \mu_sN\), where \(F_s = 4.11\) and \(\mu_s = 0.325\)

\(4.11 = (0.325)N\)

\(N = \frac{4.11}{0.325}\)

\(N = 12.65\)