Воднокруговом турнире по футболу каждая команда набрала 10 очков. сколько всего могло быть команд.

First, we'll try to plug in the value: #lim_{x to -oo}x+sqrt(x^2+2x) = -oo + sqrt(oo-oo)# we're already encountering a problem: it is simply not allowed to have #oo-oo#, it's like dividing by zero. we need to try a different approach. whenever i see this kind of limit, i try to use a trick: #lim_{x to -oo}x+sqrt(x^2+2x)# #= lim_{x to -oo}x+sqrt(x^2+2x)*(x-sqrt(x^2+2x))/(x-sqrt(x^2+2x))# these are the same becaus the factor we're multiplying with is essentially #1#. why are we doing this? because there exists a formula which says: #(a-b)(a+b) = a^2-b^2# in this case #a = x# and #b = sqrt(x^2+2x)# let's apply this formula: #lim_{x to -oo}(x^2-(sqrt(x^2+2x))^2)/(x-sqrt(x^2+2x))# #= lim_{x to -oo}(x^2-x^2-2x)/(x-sqrt(x^2+2x))# #= lim_{x to -oo}(-2x)/(x-sqrt(x^2+2x))# now we're going to use another trick. we'r going to use this one, because we want to get the #x^2# out of the square root: #lim_{x to -oo}(-2x)/(x-sqrt(x^2(1+2/x))# if you look carefully, you see it's the same thing. now, you might say that #sqrt(x^2) = x#, but you have to remember that #x# is a negative number. because we're taking the positive square root, #sqrt(x^2) = -x# in this case. #= lim_{x to -oo}(-2x)/(x+xsqrt(1+2/x))# #= lim_{x to -oo}(-2x)/(x(1+sqrt(1+2/# we can cancel the #x#: #= lim_{x to -oo}(-2)/(1+sqrt(1+2/x))# and now, we can finally plug in the value: #= -2/(1+sqrt(1+2/-oo))# a number divided by infinity, is always #0#: #= -2/(1+sqrt(1+0)) = -2/(1+1) = -2/2 = -1# this is the final answer. hope it helps.

44%

Объяснение:

25 л - 20%

40 л - 50%

Отлили 40% раствора, то есть 25*0,4=10 л 20% раствора

В 10 л 20% раствора содержится 10*0,2=2 л -соли

В 40 л 50% раствора содержится 40*0,5=20 л соли

Смесь стала 40+10=50 л, в которой содержится 20+2=22 л соли.

50 л -100%

22л - х%

х=22*100/50=44%