Recent content by mpx86

Homework Statement If angles of a quadrilateral are in the ratio 2:7:2:7, then the quadrilateral is? a) Square b) Rectangle c) Rhombus d) Parallelogram Homework Equations 2x+7x+2x+7x= 360 x=20[/B]The Attempt at a Solution Since two pairs of opposite angles of a quadrilateral are equal, hence...

lol solved... there was an error in question statement: question was 2^32 – {(2 + 1) (2^2 + 1) (2^4+1) (2^8+1) (2^16+1)} (1+x) (1+x^2) = (x^4-1)/x-1 using this iteration yields answer as 1 thanks anyways :)

Homework Statement 2^32 – (2 + 1) (2^2 – 1) (2^4+1) (2^8+1) (2^16+1)} is equal to Homework EquationsThe Attempt at a Solution Solved it by opening the bracket Answer: 2^31 + 2^24 + 2^ 18 - 2^7 + 2 Option' are 0 1 2 2^16 None of the options matched... Is there a mistake in question statement...

You could also solve it in your head using : since the weights are in the ratio 1:2 therefore 2 times decrease in 90 corresponds to 1 time increase in 84 Therefore, 84-----90 85------88 86------86(which is the desired answer)

Oops, silly me.

Homework Statement A watch gains 3 minutes every hour. If it is set right at 11 a.m. on February 21st, 2012, when will the hour hand of this defective watch and a correct watch be at the same position? A) 9 p.m. on February 21st, 2012 B) 11 p.m. on February 28th, 2012 C) 11 a.m. on March 2nd...

Homework Statement A functions is defined as f(x) = ax2 + bx + c, where a, b, c are real numbers. If f(3) = f(– 2) = 0, what is the value of f(0)?Homework EquationsThe Attempt at a Solution As function is 0 at 3, -2, therefore 9a + 3b + c=0 also, 4a -2 b + c=0 c=-6a or c= 6b f(0)=c= 6a...

PURPOSE to find a function such that when it is increased by h(Constant) at x it becomes x times it value at x E and D E is shift operator such that Ef(x)=f(x+h) , E^2 f(x)=f(x+2h) and so on s.t. E^nf(x)=f(x+nh) D is differtial operator that is d/dx... STEP 4 using taylors series e^hD= E i...

f(x+h)/f(x)=x only when h tends to zero however shouldn't it work for any value of h as it wasnt assumed that h was zero in the first step ... where is the problem?

why does it work only when h tends to zero? \hat{}\[\begin{array}{l} f(x + h) = xf(x)\\ Ef(x) = xf(x)\\ E = x\\ \ln E = hD\\ \ln x = hD\\ f(x) = y\\ y\ln x = hDy\\ y\ln x = h\frac{{dy}}{{dx}}\\ \int {} \ln xdx = h\int {} dy/y\\ x\log x/e = h\ln y + \ln c\\ x\log x/e = h\ln y/C\\...