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PDF Editor FAQ

A x 4 = E, B ÷ 4 = E, C + 4 = E, D - 4 = E, and A + B + C + D = 100. What is the value of E?

A*4=EB/4=EC+4=ED-4=Ealso A+B+C+D=100 - equation (1)We can writeB=16AC=4A-4D=4A+4Now putting in equation (1)A+16A + (4A - 4) + (4A +4)=10025A=100therefore A=4Now putting in 4A=EThis implies E=16

If [math]4a = [/math][math]e[/math][math][/math], [math]\frac{b}{4} = [/math][math]e[/math][math][/math], [math]c + [/math][math]4[/math][math] = [/math][math]e[/math][math][/math], [math]d - [/math][math]4[/math][math] = [/math][math]e[/math][math][/math] and [math]a + b + c + d = 100[/math], then what is the value of [math]e[/math][math][/math]?

i) 4a=e <-> e=4e (in words, it says that “e is 4a”. )Now, there are too many (5 )variables - a, b, c, d e.We solve an equation by rewriting one variable in terms of others.Hereii) b/4=e <-> From (i) we can substitute what “e” is. b/4=4aand now, b=4a*4 <->b = 16a (b is in terms of a)iii) c+4=e <->(same concept as in ii.) c=e-4 <->c=4a-4 (c is in terms of a)iv)d-4=e <-> d= e+4 <-> d=4a+4 (d is in terms of a)Note, that now we have variables e, b, c and d expressed only in terms of aSo, from last expressiona+b+c+d=100, we can “replace“ each variable with what it is in terms of “a”.a+(16a)+(4a-4)+(4a+4) = 100 - AND THIS IS SIMPLE LINEAR EQUATION OF SINGLE VARIABLE25a =100 <-> a=100/25 <-> a=4v) Now, you found (a), remember that above we found what is each other variable in terms of a?e=4a <-> e=4*(4) M->e=16Answer to the question is: e=16

How do I integrate [math] \int\frac{x^4}{e^x+1}dx [/math]?

Without bound this integral is no Fun but If you will put limit from zero to infinity, it will be something fun to calculate.Otherwise it will Just be function of Logarithmic integral.[math]\int x^4/(e^x +1) dx = \int x^4(e^x -1)/(e^x-1)(e^x +1) dx;[/math]multiplying and dividing by[math]e^x -1,[/math]now Just use use some algebra,[math]\int x^4/(e^x -1) dx - \int 2x^4/ (e^2x -1) dx,[/math]First integral will reduce to[math]\zeta(5)\Gamma(5)[/math]second integral will also reduce to something like this,with substitution 2x =z,finally the integral will be,[math](15/16)\zeta(5)\Gamma(5). [/math]answer.

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