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

7 0

3 years ago

Write the equation of a line that is perpendicular to x=-6 and that passes through the point (-1,-2)

maxonik [38]

the equation would be y=-2

8 0

3 years ago

If 40 is 30% of number, find that number, I NEED THE ANSWER ASAP

Mkey [24]

Answer:

The number is $133\frac{1}{3}$

Step-by-step explanation:

If something is 30% of a whole it means that if whe divide the whole into 100 and take 30, you will get this number. In order to get the whole we need to divide it my the 40 by $\frac{3}{10}$ and so we get.....

40 ÷ $\frac{3}{10}$ = $133\frac{1}{3}$

3 0

3 years ago