Adding the two polynomials gives (2x^2 + 5x^2) + (3x - x) + (4 + 2) = 7x^2 + 2x + 6.

Adding the polynomials gives (3x^2 + x^2) + (2x - x) + (1 + 4) = 4x^2 + x + 5.

Using the properties of exponents, we combine the exponents: (3x + 2x - 4x) = x, thus the result is 2^x.

Using the properties of exponents, we combine the exponents: 2^(3x + 2x - 5x) = 2^0 = 1.

The passage highlights that understanding historical context is essential for grasping the roots and persistence of inequalities.

The examples are used to illustrate the complexity of inequalities, reinforcing the author's argument.

The vertex of a quadratic function is the point at which the function reaches its maximum or minimum value.

The x-intercepts of a function are the points where the graph crosses the x-axis, meaning the output of the function is zero at those points.

Using the property of exponents (a^m)^n = a^(m*n), we have (2^3)^2 = 2^(3*2) = 2^6.

Using the power of a product property, we have (x^2 * y^3)^(2) = x^(2*2) * y^(3*2) = x^4 * y^6.

Using the property of exponents, we have (x^3 * x^2) / x^4 = x^(3+2-4) = x^1.

Adding 5 to both sides gives 3x < 9, and dividing by 3 gives x < 3.

Solving the equations simultaneously gives x = 6 and y = 4, hence the solution set is (6, 4).

The two equations represent parallel lines, which means they do not intersect and thus have no solution.

Solving the inequality gives 2x < 4, thus x < 2.

Solving the system gives x = 2 and y = 3, thus the solution set is (2, 3).

To solve for x, add 4 to both sides to get 3x = 9, then divide by 3 to find x = 3.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_15 = 15/2 * (2*10 + 14*2) = 15/2 * (20 + 28) = 15/2 * 48 = 360.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_15 = 15/2 * (2*2 + 14*4) = 15/2 * (4 + 56) = 15/2 * 60 = 450.

The sum of the first n terms of a GP is given by S_n = a(1 - r^n) / (1 - r). Here, S_5 = 2(1 - 3^5) / (1 - 3) = 2(1 - 243) / (-2) = 242.

The sum of the roots is given by -b/a = 3/2.

The sum of the roots of a quadratic equation ax^2 + bx + c = 0 is given by -b/a. Here, it is -(-4)/2 = 2.

The sum of the roots is given by -b/a = 8/2 = 4.

The author maintains an optimistic tone, suggesting that change is possible and necessary.

Calculating each term, we have 5^0 = 1, 5^1 = 5, and 5^2 = 25. Therefore, 1 + 5 + 25 = 31.

Using the property of exponents, (5^3 * 5^2) = 5^(3+2) = 5^5. Thus, (5^5) / (5^4) = 5^(5-4) = 5^1 = 5.

5^(-2) is equal to 1/(5^2) = 1/25 = 0.04.

Using the property of exponents, we have 5^(2 + 3 - 4) = 5^1 = 5.

Using the property of exponents a^m / a^n = a^(m-n), we have 5^(x+1) / 5^(x-1) = 5^((x+1)-(x-1)) = 5^2.

log_10(0.1) = log_10(10^-1) = -1.