answer:Let phi = arccos x, so 0 leq phi leq pi and cos phi = x. Then [sin phi = sqrt{1 - x^2},] thus [arctan(sin(arccos x)) = arctan(sqrt{1 - x^2}).] Let psi = arctan(sqrt{1 - x^2}). Then [cos(arctan(sqrt{1 - x^2})) = cos psi = frac{1}{sqrt{1 + (sqrt{1 - x^2})^2}} = frac{1}{sqrt{1 + 1 - x^2}} = frac{1}{sqrt{2 - x^2}} = x.] Squaring both sides, we get: [frac{1}{2 - x^2} = x^2,] or [1 = x^2(2 - x^2).] Rewriting gives: [x^4 - 2x^2 + 1 = 0.] This simplifies to: [(x^2 - 1)^2 = 0,] which gives: [x^2 = 1.] Thus, x = 1 is the only positive solution in the interval [0,1] where arccos x and cos psi are defined. boxed{x = 1} is the solution.

answer:For (A234B67) to be divisible by (8,) the sum of its digits (A+2+3+4+B+6+7 = 22 + A + B) must be divisible by (8.) Since (A) and (B) are each a single digit, the possible sums (A+B) range from (0) to (18.) Therefore, (22 + A + B) ranges from (22) to (40.) The numbers in this range that are divisible by (8) are (24) and (32.) - If (22 + A + B = 24,) then (A + B = 2.) - If (22 + A + B = 32,) then (A + B = 10.) The possible values for (A+B) are (2) and (10.) Adding these together gives (2 + 10 = boxed{12}.)

answer:The equation of the hyperbola is given by x^{2}-4y^{2}=4. To find the standard form of the equation, we can divide both sides by 4 to get frac{x^2}{4} - frac{y^2}{1} = 1. From the standard form frac{x^2}{a^2} - frac{y^2}{b^2} = 1, we can identify a^2 = 4 and b^2 = 1. Therefore, we have a = 2 and b = 1. The distance c from the center to each focus on the x-axis for a hyperbola with these parameters can be calculated using the relationship c^2 = a^2 + b^2, which gives us c^2 = 4 + 1 = 5 quad Rightarrow quad c = sqrt{5}. Thus, the coordinates of the foci of the hyperbola are (±sqrt{5}, 0). Therefore, the correct answer is: [boxed{D: (±sqrt{5}, 0)}.]

answer:First Method: Discriminant Analysis 1. The given quadratic equation is (a x^{2} + b x + c = 0). 2. To determine the nature of the roots of the quadratic equation, we need to analyze the discriminant, (D). 3. The discriminant of a quadratic equation (ax^2 + bx + c = 0) is given by: [ D = b^2 - 4ac ] 4. We are given that (b > a + c > 0). We need to work with this information to show that the discriminant is non-negative. 5. Let's compare (D) with ((a+c)^2): [ D = b^2 - 4ac ] Given ((a+c)^2): [ (a+c)^2 = a^2 + 2ac + c^2 ] Let's calculate (D - (a+c)^2): [ D - (a+c)^2 = b^2 - 4ac - (a^2 + 2ac + c^2) = b^2 - 4ac - a^2 - 2ac - c^2 = b^2 - (a^2 + c^2 + 6ac) ] 6. Given (b > a + c), let's subtract (4ac) from both sides and also work in another way: [ D = b^2 - 4ac geq (a+c - 2c)^2 - 4ac = (a-c)^2 geq 0 ] 7. Since ((a-c)^2) is always greater than or equal to 0, this implies that (D geq 0). Thus, the quadratic equation has two real roots. Second Method: Function Analysis 1. Consider the quadratic function (f(x) = a x^{2} + b x + c). 2. Evaluate (f(x)) at (x = -1): [ f(-1) = a(-1)^2 + b(-1) + c = a - b + c ] Given (b > a + c), it follows that: [ a - b + c < 0 ] 3. Evaluate (f(x)) at (x = 1): [ f(1) = a(1)^2 + b(1) + c = a + b + c ] Given (b > a + c), it follows that: [ a + b + c > 0 ] 4. Since (f(-1) < 0) and (f(1) > 0), by the Intermediate Value Theorem, the continuous quadratic function must cross the x-axis within the interval ((-1, 1)). 5. Because a parabola (the graph of a quadratic function) only changes direction at its vertex, crossing the x-axis implies that there are two points of intersection. Therefore, the quadratic equation has two real roots. # Conclusion [ boxed{text{True}} ]