Rom

who is Rohit?

Assuming you mean Jack...

\(T = \text{time spent driving first 785 miles}\\ s_1 = \dfrac{785}{T}\\ s_1+12 = \dfrac{785}{T-10}\)

\(\text{Substitute the first equation into the second and solve for } T\\ \text{Find }s_1 \text{ using the first equation}\\ s_2=s_1+12\)

\(6701^{83} = \sum \limits_{k=0}^{83} ~\dbinom{83}{k} (6000)^{k}(701)^{83-k}\\ \text{It should be clear that all terms but 1, }k=0 \text{ are divisible by }1000\\ \text{we are then left with }\\ (701)^{83} = \sum \limits_{k=0}^{83}\dbinom{83}{k}(700)^k\\ \text{here there are two terms that are not divisible by }1000,~k=0,1\\ (701)^{83}\pmod{1000} = \\ 1+83(700) \pmod{1000} = \\ 1 + (80+3)(700) \pmod{1000}= \\ 1+ 56000+ 2100 \pmod{1000} = \\ 2101 \pmod{1000} = 101\)

\(\text{let the number be }abc\\ 49a+7b+c = 81c+9b+a\\ 48a-2b=80c\\ c = \dfrac{24a-b}{40}\)

I'll let you figure out what the valid numbers are using the equation for c above.

Remember digits can only range from 0-6.

And of course digits have to be whole numbers.

\(\dfrac{4x}{(x-5)(x-3)^2} = \dfrac{A(x-3)^2+B(x-5)(x-3)+C(x-3)}{(x-5)(x-3)^2}\)

\(4x = A(x^2-6x+9) + B(x^2-8x+15)+C(x-3)\\ 4x = x^2(A + B) + x(-6A-8B+C)+(9A+15B-3C)\)

\(A+B=0\\ (-6A-8B+C)=4\\ (9A+15B-3C) = 0\)

\(\text{I leave you to solve for }A,B,C\)

\((2^3)^n=2^1\\ 2^{3n}=2^1\\ 3n=1\\ n=\dfrac 1 3\)

\(\tan(60^\circ)=\dfrac{top}{50}\\ \tan(40^\circ)=\dfrac{base}{50}\\ top-base =50 \tan(60^\circ)-50\tan(40^\circ)\\ \text{I leave you to complete it}\)

\((3m)^2 + 12 = 3(m^2 + 12)\\ 9m^2 + 12 = 3m^2 + 36\\ 6m^2-24=0\\ m^2 - 4 = 0\\ \text{I leave you to finish}\)

\(\text{This is just a point slope problem in disguise}\\ \text{the two points are }(x_1,y_1)= (1,5),~(x_2,y_2) = (-1,1)\\ \text{determine the slope of the line } a = \dfrac{y_2-y_1}{x_2-x_1}\\ \text{use that with either point to determine }b\\ \text{then }h(6) = a (6) + b\)

it looks to be NUMBER_MAX in Java

\(\cos(315^\circ) = \dfrac{\sqrt{2}}{2}\\ \sin(315^\circ) = -\dfrac{\sqrt{2}}{2}\\ z=\dfrac{3\sqrt{2}}{2} - i \dfrac{3\sqrt{2}}{2}\\ \text{now just apply the results of the last problem}\)