Rom

\(\text{let }f(n) = 5^{2n} - 2^{3n} =25^n - 8^n\\ f(1) = 25-8 = 17 \text{ which is clearly divisible by 17}\)

\(\text{Now assume that }f(n) \text{ is divisible by 17. We need to show that }f(n+1) \text{ is as well}\\ f(n+1) = 25^{n+1} - 8^{n+1} = 25\cdot 25^n - 8\cdot 8^n = \\ 25\cdot 25^n-25\cdot 8^n + 17\cdot 8^n =\\ 25(25^n - 8^n) + 17\cdot 8^n =\\ 25 \cdot 17m + 17\cdot 8^n, (\text{ f(n) =17m by assumption} ) \\ 17(25m + 8^n) \\ \text{and this is clearly divisible by 17}\)

\(\text{suppose }f(x) \text{ is a well behaved function in that it has a Taylor series}\\ f(x) = \sum \limits_{k=0}^\infty~c_k \dfrac{x^k}{k!}\)

\(\dfrac{f(x)}{x^2} = \sum \limits_{k=0}^\infty~c_k\dfrac{x^{k-2}}{k!}=\dfrac{c_0}{x^2}+\dfrac{c_1}{x}+\dfrac{c_2}{2}+\dots \text{ function of powers of x}\)

\(\text{In order for }\lim \limits_{x\to 0}\dfrac{f(x)}{x^2} = 5 \\ c_0=0,~c_1=0~c_2 = 10 \\ \text{so } f(x) = 5x^2+\text{higher powers of x}\)

a) \(\lim \limits_{x \to 0} f(x) = c_0 = 0\)

b) \(\lim \limits_{x\to 0}\dfrac{f(x)}{x} = c_1 = 0\)

a) \(E[X]=\sum \limits_{k=1}^6~\dfrac 1 6 (6k) = \dfrac{6\cdot 7}{2} = 21\)

b) \(E[X] = \dfrac 1 6(6 + 6 + 6 - 6 + 6 - 6) = 2\) (1,2,3,5 are !composite, 4 and 6 are composite)

\(\text{l =AP will be perpendicular to OA as AP is tangent to the circle}\\ \text{the slope of OA is }m_{OA} = \dfrac{7-0}{1-0} = 7 \\ \text{The slope of AP is thus } m_{AP} = -\dfrac{1}{m_{OA}}=-\dfrac 1 7\)

\(\text{The line with slope }m_{AP} \text{ passing through point }A \text{ is given by}\\ (y-7) = -\dfrac 1 7(x-1) \\ y = \dfrac{-x + 50}{7} \\ \text{and this line intersects the x axis at} \\ 0=\dfrac{-x+50}{7} \Rightarrow x = 50\)

\(\text{so }P=(50,0) \\ |OA| = \sqrt{50} \text{ as this is the radius of the circle} \\ |AP| = \sqrt{(50-1)^2 + (0-7)^2} = \sqrt{2450} = 35\sqrt{2}\\ \text{As OAP is a right triangle, the area is given by}\\ area = \dfrac 1 2 b h = \dfrac 1 2|AP| |OA|= \dfrac 1 2 \sqrt{50}\cdot35\sqrt{2} = 175\)

just apply what I showed you before

\(pV = n R T\)

\(T_1 = \dfrac{pV}{nR} \\ T_2 = \dfrac{2pV}{nR} \\ \dfrac{T_2}{T_1} = 2\\ \text{i.e. the temperature doubles when you double the pressure}\)

I'm not sure how you are supposed to figure this out but

\(f(x) = x^4 - 2x^2 \\ \text{and is of degree 4}\)

been a long time but I think

\(p_1 v_1 = p_2 v_2 \\ 100~kPa \cdot 200~L = p_2 ~kPa \cdot 80~L \\ p_2 = 250~kPa\)

\(\text{I'm going to call your roots }(r_{11},~r_{12},~r_{21},~r_{22})\)

\(r_{21}=r_{11}-1 \\ r_{22}=r_{12}-1\)

\(p_1=x^2+c x + a = (x-r_{11})(x-r_{12}) =x^2 -\left(r_{11}+r_{12}\right) x+r_{11} r_{12}\\ p_2 = x^2 +(2-r_{11}-r_{12})x + (1-r_{11}-r_{12}+r_{11}r_{12})\)

\(\text{the coefficient }a \text{ appears in both polynomials so we can solve for the roots}\\ r_{11}r_{12} = 2-r_{11}-r_{12} \\ \text{Solving this we find that there are two possibilities}\\ (r_{11},~r_{12}) = (-2,~-4) ~\text{or}\\ (r_{11},~r_{12}) = (0,~2) \)

\(\text{In the first case we then have}\\ (r_{21},~r_{22}) = (-3,-5) \\ p_1(x) = (x+2)(x+4) = x^2 + 6x+8 \Rightarrow a=8,~c=6\\ p_2(x) = (x+3)(x+5) = x^2 + 8x + 15 \Rightarrow b=15 \\ a+b+c = 8 + 15 + 6 = 29\)

\(\text{In the second case we have}\\ (r_{21},~r_{22}) = (-1,1)\\ p_1(x) =(x-0)(x-2) = x^2 - 2x \Rightarrow a=0,~c=-2\\ p_2(x) = (x+1)(x-1) = x^2 - 1 \Rightarrow b=-1\\ a+b+c = 0-1-2 = -3\)

Someone please double check all of this.

There must be a typo here.  This polynomial can't be factored over the reals.