25) How many liters of a 40% acid solution should be mixed with 8 liters of a 25% acid solution to obtain a solution that is 30% acid? State what x represents, state the equation, and then state the answer.
Amount
40% acid
25% acid
30% acid (Mixture)
Amount
Of Acid
0340x
0.250
70+)
?
Amount of acid in 40% + Amount of acid in 25% * Amount of acid in mixture

25) How many liters of a 40% acid solution should be mixed with 8 liters of a 25% acid solution to

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Ответ:

ZillaKami

29.12.2020

Mathematics

Answer is $\displaystyle\int \frac{\sec^2\left(\frac{1}{x^9}\right)}{ x^{10}} \, dx = -\frac{1}{9}\tan \left(\frac{1}{x^9}\right)+c$ ======i assume this is what this is what question asks for as it makes the substitution work: $\displaystyle\int \frac{\sec^2\left(\frac{1}{x^9}\right)}{ x^{10}} \, dx$ if $u = \dfrac{1}{x^9} = x^{-9}$ then by simple differentiation, we have $\begin{aligned} du & = -9x^{-10} dx \\ du & = -\frac{9}{x^{10}} dx\\ -\frac{x^{10}}{9} du & = dx \\ dx & = -\frac{x^{10}}{9} du \end{aligned}$ note that $\int \sec^2 u \; du = \tan u + c$ therefore, performing these substitutions: $\begin{aligned} \int \frac{\sec^2\left(\frac{1}{x^9}\right)}{ x^{10}} \, dx & = \int \frac{\sec^2\left(u\right)}{ x^{10}} \cdot \left(-\frac{x^{10}}{9}\right) \; du \\ & = \int \sec^2\left(u\right) \cdot \left(-\frac{1}{9}\right)\; du \\ & = -\frac{1}{9}\int \sec^2\left(u\right) \; du \\ & = -\frac{1}{9}\tan u + c \\ & = -\frac{1}{9}\tan \left(\frac{1}{x^9} \right) + c \end{aligned}$

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