$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

4 0

3 years ago

I will give Brainliest!! Please answer the question by simplifying!! Thanks so much and stay safe!!

Vadim26 [7]

Distribute 5/7 through the parentheses and add like terms after.

Answer: 11/7x + 11/35y

4 0

3 years ago

Will someone please help me solve this !!

Cerrena [4.2K]

360=95+120+120+125+x+84+x

360=544+2x

-184=2x

x= -92

7 0

3 years ago

{(-4, 1), (-3,5), (-1,0), (6, 2), (9,5)} is this a function or not?

tangare [24]

Yes, it’s a function because there is no repeating x-values,

7 0

3 years ago