answer:To determine which of the given points lies on the line with a slope of 45^{circ} passing through the point (1,2), let's follow these steps:Step 1: Convert the slope into a numerical value. Since the slope of the line is 45^{circ}, we know that:[k = tan(45^{circ}) = 1]Step 2: Formulate the equation of the line. Given the slope (k = 1) and a point (1,2) through which the line passes, we can use the point-slope form of a line equation. The point-slope form is y - y_1 = m(x - x_1), where (x_1, y_1) is a point on the line and m is the slope. Substituting m = 1, x_1 = 1, and y_1 = 2, we get:[y - 2 = 1 cdot (x - 1)]Step 3: Simplify the equation. Expanding the equation, we get:[y - 2 = x - 1]Simplifying further leads to:[x - y + 1 = 0]Step 4: Verify which point lies on this line. We need to check each option by substituting the x and y values into the line equation x - y + 1 = 0.- For A: (-2, 3), we get -2 - 3 + 1 neq 0.- For B: (0, 1), substituting x = 0 and y = 1 into the equation, we get:[0 - 1 + 1 = 0]This verifies that point B lies on the line.- For C: (3, 3), substituting x = 3 and y = 3 into the equation, we get:[3 - 3 + 1 neq 0]- For D: (3, 2), substituting x = 3 and y = 2 into the equation, we get:[3 - 2 + 1 neq 0]Since only the coordinates of point B satisfy the equation of the line, the correct answer is:[boxed{B}]

answer:To solve for the probability of drawing a white ball from the box, we start by identifying the total number of balls and the number of white balls in the box. Given:- The total number of balls in the box is 5 red balls + 4 white balls = 9 balls.- The number of white balls in the box is 4.The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes. In this case, the event is drawing a white ball, and the favorable outcomes are the white balls.Therefore, the probability of drawing a white ball is calculated as:[ text{Probability} = frac{text{Number of white balls}}{text{Total number of balls}} = frac{4}{9} ]Thus, the probability of drawing a white ball from the box is boxed{frac{4}{9}}.

answer:-1.7 .Let the common difference be d. Thenbegin{array}{l}a_{n}=a_{1}+(n-1) d . text { By } S_{3}=S_{11} Rightarrow d=-frac{2}{13} a_{1}<0 . text { Hence } a_{n}=a_{1}+(n-1)left(-frac{2}{13} a_{1}right) =frac{a_{1}}{13}(15-2 n),end{array}and the largest positive integer n for which a_{n} geqslant 0 is n=7.

answer:Answer: frac{3}{4}

answer:Let u=sqrt{x^{2}-1}. Squaring both sides and substituting, we rearrange the equation as follows:x^{2}left(1-frac{1}{sqrt{x^{2}-1}}right)^{2}=left(frac{91}{60}right)^{2} ;left(u^{2}+1right)left(1-frac{1}{u}right)^{2}=left(frac{91}{60}right)^{2} ;left(u+frac{1}{u}right)left(u+frac{1}{u}-2right)=left(frac{91}{60}right)^{2}This is a quadratic equation in left(u+frac{1}{u}right), whose roots are frac{169}{60} and -frac{49}{60}. However, by definition, u and u+frac{1}{u} are positive, so the negative root cannot be considered. Therefore,u+frac{1}{u}=frac{169}{60}, quad text { i.e. } quad u^{2}-frac{169}{60} u+1=0This quadratic equation has two roots: frac{5}{12} and frac{12}{5}. If u=frac{5}{12}, then x= pm sqrt{u^{2}+1}= pm frac{13}{12}. By substitution, it can be verified that -frac{13}{12} is a solution, while +frac{13}{12} is not. If u=frac{12}{5}, then x= pm sqrt{u^{2}+1}= pm frac{13}{5}. Among these, +frac{13}{5} is a solution, while -frac{13}{5} is not.Therefore, the equation has two roots: x_{1}=-frac{13}{12} and x_{2}=frac{13}{5}.

answer:14. C