The variables x and y satisfy the equation y = Ae^(-nx²), where A and n are constants. The graph of lny against x² is a straight line passing through the points (1.64; 1.76) and (3.69; 2.32). Find the values of A and the values of n correct to 2 decimal places.

Take the natural logarithm of both sides of the equation $y = Ae^{-nx^2}$:

$$\ln y = \ln(Ae^{-nx^2})$$

Using the logarithm properties $\ln(ab) = \ln a + \ln b$ and $\ln(e^k) = k$:

$$\ln y = \ln A + \ln(e^{-nx^2})$$ $$\ln y = \ln A - nx^2$$

Rearrange this equation to match the form of a straight line $Y = mX + c$:

$$\ln y = (-n)x^2 + \ln A$$

Here, $Y = \ln y$, $X = x^2$, the gradient $m = -n$, and the Y-intercept $c = \ln A$.

The straight line passes through the points $(X_1, Y_1) = (1.64, 1.76)$ and $(X_2, Y_2) = (3.69, 2.32)$.
The gradient $m$ is given by:

$$m = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{2.32 - 1.76}{3.69 - 1.64}$$ $$m = \frac{0.56}{2.05}$$ $$m \approx 0.27317073$$

From Step 1, we know that $m = -n$. Therefore:

$$n = -m = -\frac{0.56}{2.05}$$ $$n \approx -0.27317073$$

Rounding to 2 decimal places:

$$n \approx -0.27$$

Using the equation $Y = mX + c$ and the point $(X_1, Y_1) = (1.64, 1.76)$:

$$1.76 = \left(\frac{0.56}{2.05}\right)(1.64) + c$$ $$c = 1.76 - \frac{0.56 \times 1.64}{2.05}$$ $$c = 1.76 - \frac{0.9184}{2.05}$$ $$c = \frac{1.76 \times 2.05 - 0.9184}{2.05}$$ $$c = \frac{3.608 - 0.9184}{2.05}$$ $$c = \frac{2.6896}{2.05} \approx 1.3119999$$

From Step 1, we know that $c = \ln A$. Therefore:

$$A = e^c$$ $$A = e^{\frac{2.6896}{2.05}}$$ $$A \approx e^{1.3119999} \approx 3.71320$$

Rounding to 2 decimal places:

$$A \approx 3.71$$

The values are:

$$A = 3.71, n = -0.27$$