answer:The limit of f(x) does not exist in this case because the function's magnitude becomes unbounded as x approaches a. The definition of a limit requires that the function's value approaches a specific finite number. However, if q(x) to 0 while p(x) remains non-zero, the fraction frac{p(x)}{q(x)} either becomes very large or very small, resulting in no finite limit. We can prove this by contradiction. Assume that lim_{x to a} p(x) = b neq 0 and lim_{x to a} q(x) = 0. If lim_{x to a} frac{p(x)}{q(x)} = L existed, then by the product law of limits: lim_{x to a} p(x) = lim_{x to a} left(frac{p(x)}{q(x)} cdot q(x)right) = lim_{x to a} frac{p(x)}{q(x)} cdot lim_{x to a} q(x) = L cdot 0 = 0 This contradicts the assumption that lim_{x to a} p(x) neq 0. Illustratively, consider the behavior of frac{x}{1} and frac{1}{x} as x approaches zero. As x becomes smaller (e.g., x = 0.001, 0.00001), frac{x}{1} approaches zero, while frac{1}{x} grows without bound. The divergence in the latter case demonstrates the lack of a limiting value. In summary, the limit does not exist when only the denominator vanishes, as the function's behavior becomes unbounded, whereas if only the numerator vanishes, the result tends to zero.

answer:To solve the equation 9 = sqrt(3x), we need to isolate x on one side of the equation. First, we can square both sides of the equation to get rid of the square root: (9)^2 = (sqrt(3x))^2 81 = 3x Next, we can divide both sides of the equation by 3 to solve for x: 81 / 3 = 3x / 3 27 = x Therefore, the solution to the equation 9 = sqrt(3x) is x = 27. x = 27

answer:Given the scenario where Jare tosses a coin 10 times, with each heads resulting in a 1-block walk north and each tails resulting in a 1-block walk south, we need to find the probability that he finishes 4 blocks south of his original position. Let X represent the net number of blocks walked north. The experiment follows a binomial distribution with parameters n = 10 (the number of trials) and p = 1/2 (the probability of success, either heads or tails). To end 4 blocks south, Jare must have 3 heads (3 blocks north) and 7 tails (7 blocks south). The specific sequence would be HHTTTTTTTT. The number of ways this can occur can be calculated using combinations: {eq}{}^{10}{C_3} = frac{10!}{3!(10-3)!} = 120 {/eq} ways. The probability of a specific sequence happening is: {eq}{left( {frac{1}{2}} right)^3} times {left( {frac{1}{2}} right)^7} = left( {frac{1}{{16}}} right) times left( {frac{1}{{128}}} right) = frac{1}{{2048}} {/eq}. Since there are 120 possible sequences that result in 4 blocks south, we multiply the probability of one sequence by the number of sequences: {eq}120 times frac{1}{{2048}} = frac{{120}}{{2048}} = 0.05859375 {/eq}. Therefore, the probability that Jare ends up 4 blocks south is approximately 0.0586 or 5.86%.

answer:The addition of the two matrices is: left( begin{array}{cccc} -frac{12}{5} + left(-frac{17}{10}right) & -frac{1}{2} + 1 & frac{63}{10} - frac{7}{2} & frac{15}{2} - frac{9}{2} frac{51}{10} - frac{34}{5} & -frac{61}{10} - frac{34}{5} & frac{28}{5} - frac{11}{10} & -10 - frac{71}{10} end{array} right) left( begin{array}{cccc} -frac{24}{10} - frac{17}{10} & -frac{1}{2} + frac{2}{2} & frac{126}{20} - frac{35}{20} & frac{15}{2} - frac{9}{2} frac{102}{10} - frac{68}{10} & -frac{61}{10} - frac{68}{10} & frac{56}{10} - frac{11}{10} & -frac{100}{10} - frac{71}{10} end{array} right) left( begin{array}{cccc} -frac{41}{10} & frac{1}{2} & frac{91}{20} & frac{6}{2} frac{34}{10} & -frac{129}{10} & frac{45}{10} & -frac{171}{10} end{array} right) Simplifying further, we get: left( begin{array}{cccc} -frac{41}{10} & frac{1}{2} & frac{45.5}{10} & 3 frac{17}{5} & -frac{129}{10} & frac{9}{2} & -frac{171}{10} end{array} right) Which can be written as: left( begin{array}{cccc} -frac{41}{10} & frac{1}{2} & frac{91}{20} & 3 frac{34}{10} & -frac{129}{10} & frac{45}{10} & -frac{171}{10} end{array} right) The final simplified form is: left( begin{array}{cccc} -frac{41}{10} & frac{1}{2} & frac{14}{5} & 3 -frac{17}{10} & -frac{129}{10} & frac{9}{2} & -frac{171}{10} end{array} right)