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7

When distance is fixed and time is variable, it is often more natural to work with the reciprocal of speed, which I'll call "slowness" and denote $w=1/v$. I'll state a few easily verified properties of slowness: The time taken for a journey from $0$ to $X$ is $\int_0^X w\,\mathrm dx$. The average speed $\bar v$ over a journey is the reciprocal of the ...


3

We have, $$\sum_{\mathbf{x}\in\mathcal{S}}\exp\left[f(n)\sum_{i=1}^nx_i\right] = \sum_{\mathbf{x}\in\mathcal{S}} \prod_{(x_i)_{i=1}^{n} = \mathbf{x}} e^{f(n)x_i}=\left(e^{f(n)}+e^{-f(n)}\right)^n$$ The identity is a consequence of: $\displaystyle \left(t^{1}+t^{-1}\right)^{n} = \sum\limits_{\substack{1 \le i \le n\\ \epsilon_i = \pm1}} ...


3

We have that there are $n\choose r$ vectors in $S$ with exactly $r$ negative components, and that the sum of the components of any one of these vectors is $n-2r$, so your sum is $$A_{f(n)}(n)=\frac{1}{2^n}\sum_{\mathbf{x}\in\mathcal{S}}\exp\left[f(n)\sum_{i=1}^nx_i\right] = \frac{1}{2^n}\sum_{r=0}^n {n \choose r} \exp(f(n)(n-2r))$$ $$\le ...


3

You solved for $x$ wrong. Here is for $x$ correctly: We first multiply by $7$ to both sides: $$\frac{24+x}{7}=5,$$ then subtract $24$ to both sides: $$24+x=35,$$ therefore $x$ is left alone and we get: $$x=11.$$


2

As asked, the question is ambiguous and admits multiple answers. Here is why: Assume we start at $x=0$ at $t=0$ and then our position is described by $x(t)$ at all times. We then assume that we end up at $X_T$ at time $T$. As a reminder the definition of speed is $v(t)=\frac{dx}{dt}$. Ambiguity 1): When you say average speed is 50, you need to specify if ...


2

The basic idea is if you want the least value for $a$ you have to get the greatest value for $b$,...,$e$, so $b=c=d=e=110$. Then the average is $$\frac{a+4\cdot 110}{5}= 100$$. Now you can solve this for $a$.


1

Because the arithmetic mean is $100$, the sum of the five numbers must be $500$. The largest that $b,c,d$, and $e$ can be is $110$, so they can contribute at most $440$ to the total; that leaves a minimum of $500-440=60$ that must come from $a$.


1

First calculate $t$ (meeting time) from following equation : $1000+v_A \cdot t =v_B \cdot t$ where $v_A$ and $v_B$ are speeds of cars $A$ and $B$ , then , find meeting point $l$ from : $l=v_B \cdot t$


1

Assuming they move in same direction. Their relative separation when B starts moving is $1000$ km. Their relative velocity $=250-100 =150$ km. Using, $s_r=s_{\circ}+ut+\frac{1}{2}at^2$ $0=1000+150t+\frac{1}{2}\cdot0\cdot t^2$ $t=\dfrac{1000}{150}=6.6667$ Total distance $=s_{\circ}+u\cdot t$ $=1000+100\cdot6.6667=1666.67$


1

I feel the same way: in general, a problem in which the parameters are treated symmetrically should have a result that uses them symmetrically. In this case, I have used $\dfrac{|a-b|}{\max(a, b, 1)} $. It's not differentiable, but it does handle the switch between large and small $a$ and $b$ nicely, and it is symmetric in them.


1

Lets look at it like you need a total of $60$ percentage points. Right now you have $$.05(60)+.05(53)+.3(47)+.1(66)=26.35\text{ percentage points}$$ So, you need to now achieve $$.05(\text{A3})+.05(\text{A4})+.3(\text{T2})+.10(\text{P2})$$ such that the sum of that expression is $33.65$. If you are looking for what single grade could you score on them ...


1

Here is my attempt at trying to formulate your question. Let $d_i$ represent the distance travelled in segment $i$ and $s_i$ represent the speed in segment $i$. Then total time taken ($T$) is given by:$$T=\sum_{i=1}^n\frac{d_i}{s_i}\tag{1}$$Now, if we increase the speed in each segment by some factor $f$ ($\gt1$) then the new time taken ($T_f$) will be ...


1

$$\text{There are a lot of ways to define the minimum for the grades as a whole.}$$ $$\text{One way to define the magnitude for a set of values is distance from the origin.}$$ $$$$ $$\text{Use Lagrange Multipliers to compute the extreme values.}$$ $\text{Let } f(x, y, z, u) = x^2 + y^2 + z^2 + u^2$ $\text{Let } g(x, y, z, u) = 0.25x + 0.2y + 0.25z + 0.3u - ...


1

There are several ways to solve this. However, I've provided a comprehensive walkthrough that I think will help you understand why the methods other people will list actually work. $$\frac{a+b+c+d+e+f}{6}=4$$ $$a+b+c+d+e+f=24$$ $$\frac{24+g}{7}=5$$ $$24+g=35$$ $$g=35-24$$ $$g=11$$



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