If the coefficient of x^2 is negative, the graph will be n shaped and curve down instead of like a u shape if it was positive. If the vertex is below the x axis and curves down, it won’t pass the x axis, Tia is right.

5 0

3 years ago

Solve the differential equation.<br><br> (y^2 + xy^2)y' = 1

stira [4]

$(y^2 + xy^2)y' = 1 \ \Rightarrow\ (y^2 + xy^2)\frac{dy}{dx}= 1 \ \Rightarrow\\ \\
y^2(1 + x) \frac{dy}{dx} = 1 \ \Rightarrow\ y^2 dx = \frac{dx}{1 + x}\ \Rightarrow \\ \\
\int y^2 dx =\int \frac{dx}{1 + x} \ \Rightarrow\ \frac{1}{3}y^3 = \ln|1+x| + C\ \Rightarrow \\ \\
y^3 = 3 \ln|1 + x| + 3C\ \Rightarrow\ y = \sqrt[3]{3\ln|1+x| + K},\ \text{where $K = 3C$}
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6 0

3 years ago

When a data set is normally distributed, about how much of the data fall within one standard deviation of the mean?

Bezzdna [24]

That would be 68% answer

6 0

4 years ago