3(4 + 5) = 3 × 9 = 27.

Substituting x = 2: 3(2) + 5 = 6 + 5 = 11.

Calculating gives 3^3 = 27 and 4^2 = 16. Therefore, 27 + 16 = 43.

Calculating 3^3 gives 27 and 2^3 gives 8. Therefore, 27 - 8 = 19.

3^3 = 3 × 3 × 3 = 27.

Calculating gives 4 × 5 - 5 = 20 - 5 = 15.

According to the order of operations, we first perform the multiplication: 2 × 3 = 6. Then we add: 5 + 6 = 11.

5! = 5 × 4 × 3 × 2 × 1 = 120.

5! = 120, 3! = 6, and 2! = 2. Thus, 120 / (6 * 2) = 10.

First, calculate inside the parentheses: 2 + 3 = 5. Then, 5 × 5 - 4 = 25 - 4 = 21.

5(3 + 2) - 4 = 5 × 5 - 4 = 25 - 4 = 21.

According to the order of operations, we first calculate 3 × 2 = 6, then add 7, giving 7 + 6 = 13.

The gas constant R is 0.0821 L·atm/(K·mol).

The gravitational constant (G) is 6.67 x 10^-11 N m²/kg².

The gravitational constant G is approximately 6.67 x 10^-11 N(m/kg)^2.

Integral = [x^3 + 2x] from 0 to 1 = (1 + 2) - (0) = 3.

The integral evaluates to [x^3] from 0 to 1 = 1^3 - 0^3 = 1.

The integral ∫(0 to 1) x^2 dx = [x^3/3] from 0 to 1 = 1/3.

The integral evaluates to [x^3 - 2x] from 1 to 2 = (8 - 4) - (1 - 2) = 3.

The integral evaluates to 7.

The integral evaluates to tan^-1(x) + C.

The integral ∫(2x + 3)dx = [x^2 + 3x] from 0 to 1 = (1^2 + 3*1) - (0 + 0) = 1 + 3 = 4.

∫(2x)dx = x^2 from 0 to 1 = 1^2 - 0^2 = 1.

∫(2x)dx = x^2 from 0 to 2 = 2^2 - 0^2 = 4.

The integral evaluates to [0.5x^4 - x^3 + 4x] from 1 to 2 = (8 - 8 + 8) - (0.5 - 1 + 4) = 8 - 2.5 = 5.5.

The integral evaluates to 20.

Evaluating the integral gives [x^3 + x^2]_0^1 = (1 + 1) - (0 + 0) = 2.

Using L'Hôpital's rule, the limit evaluates to 5.

Using L'Hôpital's Rule, the limit evaluates to 2.

The limit lim(x->0) (sin(x)/x) = 1.