Solve the math problems in the image.

Steps to Solve

1. Find the general term of the arithmetic sequence

The sequence is 2, 5, 8, 11,... The common difference $d$ is $5-2 = 3$. The first term $a_1$ is 2. The general term of an arithmetic sequence is given by $a_n = a_1 + (n-1)d$.

2. Calculate the 2001st term

Substitute $n=2001$, $a_1 = 2$, and $d = 3$ into the formula for the nth term. $a_{2001} = 2 + (2001-1)(3) = 2 + (2000)(3) = 2 + 6000 = 6002$.

3. Calculate the area of the right triangle using the legs

The legs of the right triangle are AC = 6 and BC = 8. The area of the triangle is given by $\frac{1}{2} \times base \times height = \frac{1}{2} \times 6 \times 8 = 24$ square units.

4. Calculate the perpendicular distance from C to AB

Let the perpendicular distance from C to AB be $h$. The area of the triangle can also be calculated as $\frac{1}{2} \times AB \times h$. Since AB = 10, we have $\frac{1}{2} \times 10 \times h = 5h$. Equating the two area calculations, we have $5h = 24$, so $h = \frac{24}{5} = 4.8$ units.

5. Simplify the expression with exponents

We have to simplify $\frac{2^{2001} + 2^{1999}}{2^{2000} - 2^{1998}}$. We can factor out common terms in the numerator and the denominator. Numerator: $2^{2001} + 2^{1999} = 2^{1999}(2^2 + 1) = 2^{1999}(4 + 1) = 5 \cdot 2^{1999}$ Denominator: $2^{2000} - 2^{1998} = 2^{1998}(2^2 - 1) = 2^{1998}(4 - 1) = 3 \cdot 2^{1998}$ Therefore, $\frac{2^{2001} + 2^{1999}}{2^{2000} - 2^{1998}} = \frac{5 \cdot 2^{1999}}{3 \cdot 2^{1998}} = \frac{5}{3} \cdot 2^{1999-1998} = \frac{5}{3} \cdot 2^1 = \frac{10}{3}$.

6. Calculate $x^4$ given $x^8 = 64$

We are given $x^8 = 64$. We want to find $x^4$. Since $(x^4)^2 = x^8$, we can take the square root of both sides of the given equation: $\sqrt{x^8} = \sqrt{64}$ which gives $x^4 = 8$.