Find the point on x axis which is equidistant from (1; 2) and (-3; 4)

Solved

Show answers

Ответ:

mjasmine3280

01.08.2020

Mathematics

$V = \sqrt[4]{K1*t+K2}$

Step-by-step explanation:

Since:

V: volume

t: time

K: proporcional constant

dV/dt = K*V^3 ---> (V^-3)dV = Kdt

Integrating by parts: $\int\limits^{V}_{Vo} V^{-3} \, dV =K* \int\limits^{t}_{to} {t} \, dt$

Then:

$(-1/4)*(V^{-4} -Vo^{-4})= K*(t-to)$

$(-1/4)*V^{-4}= K*(t-to) + (-1/4)*Vo^{-4}$

$V^{-4}= (-4)*K*(t-to) + (-4)*(-1/4)*Vo^{-4} = -4K*(t-to) + Vo^{-4}$

If we call: K1 = -4K ; K2 = Vo^-4, (both constants) then:

$V = \sqrt[4]{K1*(t-to)+K2}$

If chose select the initial time into = 0, then:

$V = \sqrt[4]{K1*t+K2}$

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Mathematics

Ask an AI advisor a question

I want advice