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 $\frac{3}{7} + 1\frac{5}{7} \div \frac{4}{7} of 2\frac{1}{3}$. 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): $\frac{4}{7} of 2\frac{1}{3} = \frac{4}{7} \times \frac{7}{3} = \frac{4}{3}$ 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.

a) Determine the gradient of the line. Step 1: Identify the coordinates of the points. Let $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (3, 7)$.

Step 2: Use the gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. $m = \frac{7 - 3}{3 - 1}$ $m = \frac{4}{2}$ $m = 2$ The gradient of the line is $\boxed{2}$.

b) Determine the equation of the line. Step 1: Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$. We have the gradient $m=2$ and can use point A $(1, 3)$. $y - 3 = 2(x - 1)$

Step 2: Simplify the equation to the slope-intercept form $y = mx + c$. $y - 3 = 2x - 2$ $y = 2x - 2 + 3$ $y = 2x + 1$ The equation of the line is $\boxed{y = 2x + 1}$.