Solve for x in 16^x+1 = 0.25

Step 1: Express both sides of the equation with the same base.
We know that $16 = 2^4$ and $0.25 = \frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2}$.
Substitute these into the equation:

$$(2^4)^{2x+1} = 2^{-2}$$

Step 2: Apply the exponent rule $(a^m)^n = a^{mn}$.

$$2^{4(2x+1)} = 2^{-2}$$ $$2^{8x+4} = 2^{-2}$$

Step 3: Equate the exponents since the bases are the same.

$$8x+4 = -2$$

Step 4: Solve for $x$.

$$8x = -2 - 4$$ $$8x = -6$$ $$x = \frac{-6}{8}$$ $$x = -\frac{3}{4}$$

The solution is $\boxed{-\frac{3}{4}}$.

Step 1: Evaluate the numerator.
The numerator is $\frac{3}{7} + 1\frac{5}{7} \div \frac{4}{7} \text{ of } 2\frac{1}{3}$.
Convert mixed numbers to improper fractions:
$1\frac{5}{7} = \frac{1 \times 7 + 5}{7} = \frac{12}{7}$
$2\frac{1}{3} = \frac{2 \times 3 + 1}{3} = \frac{7}{3}$
Now, calculate "of" (multiplication):
$\frac{4}{7} \text{ of } 2\frac{1}{3} = \frac{4}{7} \times \frac{7}{3} = \frac{4}{3}$
Next, perform the division:
$\frac{12}{7} \div \frac{4}{3} = \frac{12}{7} \times \frac{3}{4} = \frac{36}{28} = \frac{9}{7}$
Finally, perform the addition:
$\frac{3}{7} + \frac{9}{7} = \frac{12}{7}$
So, the Numerator is $\frac{12}{7}$.

Step 2: Evaluate the denominator.
The denominator is $(1\frac{3}{7} - \frac{5}{8}) \times \frac{2}{3}$.
Convert the mixed number to an improper fraction:
$1\frac{3}{7} = \frac{1 \times 7 + 3}{7} = \frac{10}{7}$
Perform the subtraction inside the parenthesis:
$\frac{10}{7} - \frac{5}{8} = \frac{10 \times 8}{7 \times 8} - \frac{5 \times 7}{8 \times 7} = \frac{80}{56} - \frac{35}{56} = \frac{45}{56}$
Next, perform the multiplication:
$\frac{45}{56} \times \frac{2}{3} = \frac{45 \times 2}{56 \times 3} = \frac{90}{168}$
Simplify the fraction by dividing by common factors (e.g., 6):
$\frac{90 \div 6}{168 \div 6} = \frac{15}{28}$
So, the Denominator is $\frac{15}{28}$.

Step 3: Divide the Numerator by the Denominator.

$$\frac{\frac{12}{7}}{\frac{15}{28}} = \frac{12}{7} \div \frac{15}{28}$$ $$= \frac{12}{7} \times \frac{28}{15}$$ $$= \frac{12 \times 28}{7 \times 15}$$

Simplify by canceling common factors (7 with 28, and 3 with 12 and 15):

$$= \frac{4 \times 4}{1 \times 5}$$ $$= \frac{16}{5}$$

The value is $\boxed{\frac{16}{5}}$ or $\boxed{3.2}$.

Step 1: Simplify $(1+\sqrt{3})(1-\sqrt{3})$.
This is in the form $(a+b)(a-b) = a^2 - b^2$.
Here, $a=1$ and $b=\sqrt{3}$.

$$(1+\sqrt{3})(1-\sqrt{3}) = 1^2 - (\sqrt{3})^2$$ $$= 1 - 3$$ $$= -2$$

The simplified expression is $\boxed{-2}$.

Step 2: Evaluate $\frac{1}{1+\sqrt{3}}$ to 3 significant figures.
First, rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $(1-\sqrt{3})$.

$$\frac{1}{1+\sqrt{3}} = \frac{1}{1+\sqrt{3}} \times \frac{1-\sqrt{3}}{1-\sqrt{3}}$$ $$= \frac{1-\sqrt{3}}{(1+\sqrt{3})(1-\sqrt{3})}$$

Using the result from Step 1 for the denominator:

$$= \frac{1-\sqrt{3}}{-2}$$

Substitute the given value $\sqrt{3} = 1.7321$:

$$= \frac{1 - 1.7321}{-2}$$ $$= \frac{-0.7321}{-2}$$ $$= 0.36605$$

Round the result to 3 significant figures:
The first three significant figures are 3, 6, 6. The next digit is 0, so we round down.

$$0.366$$

The evaluated expression is $\boxed{0.366}$.

The question asks to "Determine:" for a line passing through points A (1, 3) and B (3, 7). We will determine the gradient and the equation of the line.