answer:To find ( x = g^{-1}(-7/64) ), we solve the equation: [ frac{x^5 - 1}{4} = -frac{7}{64}. ] Multiplying both sides by 4: [ x^5 - 1 = -frac{7}{16}. ] Adding 1 to both sides: [ x^5 = -frac{7}{16} + 1 = -frac{7}{16} + frac{16}{16} = frac{9}{16}. ] Taking the fifth root: [ x = left(frac{9}{16}right)^{frac{1}{5}}. ] Conclusion: [ x = boxed{left(frac{9}{16}right)^{frac{1}{5}}}. ]

answer:Since triangle XYZ is a right triangle with angle X = angle Z, it must be an isosceles right triangle (as it has two congruent acute angles). Here, overline{XZ} acts as the hypotenuse. - First, find the lengths of the legs XY and YZ which are equal due to the isosceles property. Since overline{XZ} is the hypotenuse in this right triangle, we can use the property of isosceles right triangles where the legs are frac{text{hypotenuse}}{sqrt{2}}. Hence, both XY and YZ are equal to frac{8sqrt{2}}{sqrt{2}} = 8. - Calculate the area of triangle XYZ using the formula text{Area} = frac{1}{2} times text{base} times text{height}. Here, both the base and height are the legs of the triangle, so text{Area} = frac{1}{2} times 8 times 8 = 32. boxed{32}

answer:Let the common ratio of the geometric sequence {a_n} be q. Given a_1=8 and a_4=a_3a_5, We know that in a geometric sequence, a_n = a_1 * q^{n-1}. So, a_4 = a_1 * q^{4-1} = 8q^3 a_3 = a_1 * q^{3-1} = 8q^2 a_5 = a_1 * q^{5-1} = 8q^4 Substitute these values back into the given equation a_4=a_3a_5, we get 8q^3 = 8q^2 * 8q^4 Simplify and solve for q, q^3 = 8q^6 q^3 = 2^3q^6 q^3(1-2^3q^3)=0 q^3=0 text{ or } q^3=frac{1}{2^3} Since q cannot be 0 (as it would make all terms in the sequence 0), we take the second solution, q=sqrt[3]{frac{1}{2^3}}=frac{1}{2} So, the common ratio of the sequence {a_n} is boxed{frac{1}{2}}. Thus, the answer is (A). This problem can be solved using the general term formula of a geometric sequence. It tests reasoning and computational skills and can be considered a medium-difficulty problem.

answer:To solve this problem, we start by calculating the total number of ways Person A and Person B can each choose 2 venues out of the 4 available venues at the Flower Expo. For Person A, the number of ways to choose 2 venues out of 4 is calculated using the combination formula {C}_{n}^{r} = frac{n!}{r!(n-r)!}, where n is the total number of items, and r is the number of items to choose. Therefore, for Person A, we have: {C}_{4}^{2} = frac{4!}{2!(4-2)!} = frac{4 times 3}{2 times 1} = 6 Similarly, Person B can also choose 2 venues out of 4 in the same number of ways: {C}_{4}^{2} = 6 Thus, the total number of ways both Person A and Person B can choose their venues is: 6 times 6 = 36 Next, we calculate the number of ways in which exactly one venue is the same in their choices. First, we select the common venue, which can be done in: {C}_{4}^{1} = frac{4!}{1!(4-1)!} = 4 ways. Then, from the remaining 3 venues, we need to choose 2 venues for one person and the remaining venue automatically becomes the second choice for the other person, which can be done in: {A}_{3}^{2} = frac{3!}{(3-2)!} = 3 times 2 = 6 ways. Therefore, the total number of ways to have exactly one venue in common is: 4 times 6 = 24 Finally, the probability that exactly one venue is the same in their choices is the ratio of the number of favorable outcomes to the total number of outcomes, which is: P = frac{24}{36} = frac{2}{3} Therefore, the answer is boxed{frac{2}{3}}.