Acleus bought land for Kshs. 1,450,000. After 10 years, its value was Kshs. 3,500,000. Find the annual appreciation rate to the nearest whole number.

Given:
$P = 1,450,000$
$A = 3,500,000$
$n = 10$ years
$r = ?$

The rate of appreciation is \boxed{\text{9%}}.

The subject of the formula is $\boxed{x = \frac{P^2 d^2 - g}{1 + P^2 d^2}}$.

For similar solids, the ratio of their masses (which is proportional to volume) is equal to the cube of the ratio of their linear dimensions, and the ratio of their surface areas is equal to the square of the ratio of their linear dimensions.
Let $M_L$ be the mass of the larger container and $M_S$ be the mass of the smaller container.
Let $SA_L$ be the surface area of the larger container and $SA_S$ be the surface area of the smaller container.

Given:
$M_L = 256\text{ kg}$
$M_S = 121\text{ kg}$
$SA_S = 825\text{ cm}^2$

The ratio of masses is $\frac{M_L}{M_S} = \frac{256}{121}$.
The ratio of surface areas is $\frac{SA_L}{SA_S}$.

Since mass is proportional to volume ($L^3$) and surface area is proportional to ($L^2$), we have:

From the first equation, $\frac{L_L}{L_S} = \left(\frac{M_L}{M_S}\right)^{1/3}$.
Substitute this into the second equation:

First, calculate the cube root of the ratio:

Then square it:

Alternatively, recognize that $256 = 4^4$ and $121 = 11^2$. This doesn't simplify the cube root directly.
Let's use the property $(a/b)^{2/3} = ((a^{1/3})/(b^{1/3}))^2$.
$256^{1/3} \approx 6.3496$
$121^{1/3} \approx 4.9460$
So, $\left(\frac{256}{121}\right)^{1/3} \approx \frac{6.3496}{4.9460} \approx 1.2838$
Then, $\left(\frac{256}{121}\right)^{2/3} \approx (1.2838)^2 \approx 1.6481$

Let's re-evaluate using fractions for clarity if possible.
$256 = 4^4$
$121 = 11^2$
So, $\left(\frac{256}{121}\right)^{2/3}$ does not simplify nicely to integers. We must use decimal approximation.

Using a calculator for $\left(\frac{256}{121}\right)^{2/3}$:

The surface area of the larger container is $\boxed{1360.00\text{ cm}^2}$.

Here, $a=1$, $b=-x$, and $n=5$.

Term 1 ($k=0$): $\binom{5}{0}(1)^5(-x)^0 = 1 \times 1 \times 1 = 1$
Term 2 ($k=1$): $\binom{5}{1}(1)^4(-x)^1 = 5 \times 1 \times (-x) = -5x$
Term 3 ($k=2$): $\binom{5}{2}(1)^3(-x)^2 = 10 \times 1 \times (x^2) = 10x^2$
Term 4 ($k=3$): $\binom{5}{3}(1)^2(-x)^3 = 10 \times 1 \times (-x^3) = -10x^3$

The expansion of $(1-x)^5$ up to the term in $x^3$ is $\boxed{1 - 5x + 10x^2 - 10x^3}$.