answer:To solve this problem, we start by calculating the total number of baby grasshoppers Jemma found. Since a dozen equals 12, and she found 2 dozen baby grasshoppers, we calculate the total number of baby grasshoppers as follows: [2 times 12 = 24] Next, we add the 7 grasshoppers Jemma saw on her African daisy plant to the 24 baby grasshoppers she found hopping on the grass under the plant to find the total number of grasshoppers: [24 + 7 = 31] Therefore, the total number of grasshoppers Jemma found altogether is boxed{31}.

answer:First, let's calculate how many candies Yuna ate in a week. Since she ate 6 pieces every day for 7 days, she ate: 6 pieces/day * 7 days = 42 pieces Now, let's find out how many candies she has left after eating 42 pieces: 60 candies - 42 candies = 18 candies Yuna plans to eat 3 pieces a day from now on. To find out how many more days she can eat candy, we divide the remaining candies by the number of pieces she plans to eat per day: 18 candies / 3 pieces/day = 6 days Yuna can eat candy for boxed{6} more days.

answer:Let P(m, sqrt{m}), The derivative of the function y = f(x) = sqrt{x} is f'(x) = frac{1}{2sqrt{x}}, Thus, the slope of the tangent line at point P is k = f'(m) = frac{1}{2sqrt{m}}, The tangent line equation is y - sqrt{m} = frac{1}{2sqrt{m}}(x - m), Since the tangent line passes through F(-1, 0), - sqrt{m} = frac{1}{2sqrt{m}}(-1-m), This simplifies to 2m = 1 + m, which yields m = 1, so P(1, 1), As the left focus is F(-1, 0), the right focus is F_1(1, 0), Thus, c = 1, and 2a = |PF| - |PF_1| = sqrt{(-1-1)^2 + 1} - 1 = sqrt{5} - 1, This gives a = frac{sqrt{5} - 1}{2}, The eccentricity of the hyperbola is e = frac{c}{a} = frac{1}{frac{sqrt{5} - 1}{2}} = frac{2}{sqrt{5} - 1} cdot frac{2(sqrt{5} + 1)}{5 - 1} = boxed{frac{sqrt{5} + 1}{2}}, Therefore, the answer is (C). To solve this problem, we assumed the coordinates of point P, found the derivative of the given function, used the geometric meaning of the derivative to find the tangent line equation, determined the coordinates of P, and used the definition of a hyperbola to find a. This allowed us to calculate the eccentricity. The main focus of this problem is calculating the eccentricity of a hyperbola, and the key to solving it lies in determining the slope and equation of the tangent line using the geometric meaning of the function's derivative.

answer:We are given a circle in which an angle with sides of lengths 1 and 2 is inscribed and subtends an arc of 120^circ. We need to find the radius of this circle. 1. **Identify the given values and convert angle measurements**: We are given that: [ AC = 1, quad AB = 2, quad text{and the arc is } 120^circ. ] According to the inscribed angle theorem, the angle angle BAC is half the measure of the arc it subtends: [ angle BAC = frac{120^circ}{2} = 60^circ. ] 2. **Use the Law of Cosines**: To find the length of segment BC using the Law of Cosines, we have: [ BC = sqrt{AC^2 + AB^2 - 2 cdot AC cdot AB cdot cos (angle BAC)} ] Substitute the known values: [ BC = sqrt{1^2 + 2^2 - 2 cdot 1 cdot 2 cdot cos (60^circ)} ] Since cos (60^circ) = frac{1}{2}: [ BC = sqrt{1 + 4 - 2 cdot 1 cdot 2 cdot frac{1}{2}} ] Simplify the expression inside the square root: [ BC = sqrt{1 + 4 - 2} = sqrt{3} ] 3. **Use the Law of Sines to find the radius R**: The Law of Sines states: [ frac{BC}{sin (angle BAC)} = 2R ] We know angle BAC = 60^circ and sin (60^circ) = frac{sqrt{3}}{2}. Therefore: [ R = frac{BC}{2 cdot sin (60^circ)} ] Substitute the known values: [ R = frac{sqrt{3}}{2 cdot frac{sqrt{3}}{2}} = frac{sqrt{3}}{sqrt{3}} = 1 ] # Conclusion: [ boxed{1} ]