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The brakes reduce it's velocity by $10$ m/s in every second. So the acceleration is $-10 \, \text{m/s}^2$. So $$y(t)=v-10\,\text{ms}^{-2}t$$ $$x(t)=100\,\text{ms}^{-1}t+\frac{1}{2}(-10\,\text{ms}^{-2})t^2$$


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In general $$x\in f^{-1}(X)\iff f(x)\in X\tag1$$ Observe that $x\in X\implies f(x)\in f(X)\implies x\in f^{-1}(f(X))$ so it is always true that: $$X\subseteq f^{-1}(f(X))\tag2$$ Essential is the information $f^{-1}(f(X))\subseteq X$ for every $X\subseteq A$. Applying this on $X=\{a\}$ where $a\in A$ we get $f^{-1}\left(f\left(\left\{ a\right\} \right)\...


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The ratio test will prove that the series is absolutely convergent if the absolute value of the result is less than $1$. This is true because the series will eventually grow smaller than an arbitrary convergent geometric series. Therefore, since we know that series a and b converges with $|x|<r_a,r_b$ respectively, we also know that as $lim_{k>\infty}$,...


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Since $f_y=2y$ and $f_x=p^{\prime}(x)$, you need to find $p(x)$ so that $p^{\prime}(x)=0$ for 3 values of $x$, and since $D=2p^{\prime\prime}(x)$, you also need $p^{\prime\prime}(x)<0$ at two of the critical points and $p^{\prime\prime}(x)>0$ at the other critical point.


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We want there to be three appropriate solutions to $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 2y = 0$$ Consequently, we will have $y = 0$ and so $$\frac{\partial f}{\partial x} = \frac{dp}{dx} = 0$$ at each critical point. Since there are three critical points of $p$, it will have a degree of at least 4. Lastly, we need to ensure that ...


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The summand is equivalent to: $\frac{1}{m} \cdot [\frac{1}{n}-\frac{1}{m+n}]$ Assuming finite integer $m$, we can begin telescoping when we are at the $m+1$th term. Hence this is just an expression for the $m$th harmonic number. (If you are unfamiliar with telescoping series: https://en.m.wikipedia.org/wiki/Telescoping_series)



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