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Y=(4*sqrt(x)+3)*(1-1/x) найти производную

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Ответы

amarantmetall
y'=(4 \sqrt{x} +3)'(1- \frac{1}{x})+(4 \sqrt{x} +3)((1- \frac{1}{x} )'\\y'= \frac{2}{ \sqrt{x} } (1- \frac{1}{x} )=(4 \sqrt{x} +3)* \frac{1}{x^2}\\y'= \frac{2x \sqrt{x} +2 \sqrt{x} +3}{x^2}
a-zotova

17 км/ч; 2,5 км/ч.

Объяснение:

Обозначим собственную скорость баржи w км/ч, а скорость течения v км/ч.

Тогда скорость по течению будет w+v км/ч, а скорость против течения w-v км/ч.

Составляем систему:

{ 6(w+v) + 4(w-v) = 175

{ 3,5(w-v) = 2,5(w+v) + 2

Второе уравнение умножаем на 2 и раскрываем скобки в обоих уравнениях.

{ 6w + 6v + 4w - 4v = 175

{ 7w - 7v = 5w + 5v + 4

Приводим подобные

{ 10w + 2v = 175

{ 2w = 12v + 4

Делим на 2 оба уравнения

{ 5w + v = 87,5

{ w = 6v + 2

Подставляем второе уравнение в первое уравнение

5(6v + 2) + v = 87,5

30v + 10 + v = 87,5

31v = 77,5

v = 77,5/31 = 2,5 км/ч - это скорость течения.

w = 6v + 2 = 6*2,5 + 2 = 17 км/ч - это собственная скорость баржи.

Vasilevich
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