3 years ago

Unit Checkpoint 3 - Grade 6 - K12

Mathematics

2 answers:

eduard3 years ago

8 0

What he said, he’s basically right my guy

Juliette [100K]3 years ago

5 0

Answer:

6, 24, 36

Step-by-step explanation:

because the ratio is 1:3

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Solve the differential equation dy dx equals the quotient of x squared and y squared for y = f(x) with the initial condition y(0

Gnoma [55]

$\displaystyle\frac{dy}{dx} = \frac{x^2}{y^2}\ \Rightarrow\ y^2 dy = x^2 dx\ \Rightarrow\ \int y^2 dy = \int x^2 dx\ \Rightarrow\textstyle\ \frac{1}{3}y^3 = \frac{1}{3}x^3 + C. \\ \\
\text{Now } y(0) = 2\ \Rightarrow\ \frac{1}{3}(2)^3 = \frac{1}{3}(0)^3 + C\ \Rightarrow\ \frac{8}{3} = C,\text{ so } \frac{1}{3}y^3 = \frac{1}{3}x^3 + \frac{8}{3}. \\ \\
y^3 = x^3 + 8\ \Rightarrow\ y = \sqrt[3]{x^3 + 8}$