answer:1. Given the problem involves two quadratic equations: [ x^2 + x + q_1 = 0 ] and [ x^2 + p x + q_2 = 0, ] where (p, q_1,) and (q_2) are real numbers, and (p = q_1 + q_2 + 1). We need to show that at least one of these equations has two distinct real roots. 2. First, analyze the discriminant of the first equation: [ Delta_1 = 1 - 4q_1. ] For this quadratic equation to not have two distinct real roots, we must have (Delta_1 leq 0), which translates into: [ 1 - 4q_1 leq 0 Rightarrow q_1 geq frac{1}{4}. ] 3. Now, consider the second quadratic equation. Substitute (p = q_1 + q_2 + 1) into the discriminant of the second equation: [ Delta_2 = p^2 - 4q_2 = (q_1 + q_2 + 1)^2 - 4q_2. ] Expand and simplify this expression: [ begin{aligned} Delta_2 &= (q_1 + q_2 + 1)^2 - 4q_2 &= q_2^2 + 2(q_1 + 1)q_2 + (q_1 + 1)^2 - 4q_2 &= q_2^2 + 2(q_1 - 1)q_2 + (1 + q_1)^2. end{aligned} ] 4. Now, examine the discriminant of the quadratic expression in (q_2): [ Delta_3 = 4(q_1 - 1)^2 - 4(1 + q_1)^2. ] Simplify (Delta_3): [ begin{aligned} Delta_3 &= 4(q_1 - 1)^2 - 4(1 + q_1)^2 &= 4[q_1^2 - 2q_1 + 1 - (q_1^2 + 2q_1 + 1)] &= 4[q_1^2 - 2q_1 + 1 - q_1^2 - 2q_1 - 1] &= 4(-4q_1) &= -16q_1. end{aligned} ] 5. Given that (q_1 geq frac{1}{4}), it follows that: [ -16q_1 leq -4. ] Therefore, (Delta_3 < 0). 6. Since (Delta_3 < 0), the expression (q_2^2 + 2(q_1 - 1)q_2 + (1 + q_1)^2) is always positive. Hence, (Delta_2 > 0), implying that the second quadratic equation (x^2 + px + q_2 = 0) has two distinct real roots. # Conclusion: At least one of the given quadratic equations has two distinct real roots. [ boxed{ } ]

answer:(1) Since the terminal side of α passes through point Pleft( frac {4}{5},- frac {3}{5}right), the radius r is obtained by the distance formula r= sqrt {left( frac {4}{5}right)^{2}+left(- frac {3}{5}right)^{2}}=1. Therefore, the sin α can be found by considering the y-coordinate of point P in the unit circle, which gives sin α = - frac {3}{5}. So the answer for the first part is boxed{ - frac {3}{5} }. (2) Knowing from (1), angle α is in the fourth quadrant, we can deduce that cos α is positive in this quadrant. So, cos α = frac {4}{5}. Using the given trigonometric identities and properties, we can rewrite the expression as follows: frac {sin left( frac {π}{2}-αright)}{sin (π+α)}cdot frac {tan (α-π)}{cos (3π-α)} = frac {cos α}{-sin α}cdot frac {tan α}{-cos α}. As tan α = frac {sin α}{cos α}, substituting tan α in the equation simplifies the expression: frac {cos α}{-sin α}cdot frac {sin α/cos α}{-cos α} = frac {1}{cos α} Since cos α = frac {4}{5}, the final value is given by: frac {1}{cos α} = frac {1}{frac {4}{5}} = frac {5}{4} So the answer to the second part is boxed{frac {5}{4}}.

answer:Let the original cost of the article be Rs. x. After the first discount of 20%, the cost of the article becomes: x - (20/100) * x = 0.8 * x After the second discount of 24% on the reduced price, the cost of the article becomes: 0.8 * x - (24/100) * (0.8 * x) = 0.76 * (0.8 * x) = 0.608 * x We are given that after both discounts, the final cost of the article is Rs. 320. Therefore, we can set up the equation: 0.608 * x = 320 Now, we can solve for x: x = 320 / 0.608 x = 526.31578947 The original cost of the article was approximately Rs. boxed{526.32} .

answer:Let's call the certain percentage "p". We are given that 50% of x is 20 less than p% of 500, and that x is 10. First, let's find 50% of x: 50% of x = 0.5 * x Since x is 10: 0.5 * 10 = 5 Now, we know that 5 is 20 less than p% of 500. Let's set up an equation to find p: p% of 500 - 20 = 5 To express p% as a decimal, we divide p by 100: (p/100) * 500 - 20 = 5 Now, let's solve for p: (p/100) * 500 = 5 + 20 (p/100) * 500 = 25 Now, we multiply both sides by 100 to get rid of the fraction: p * 500 = 25 * 100 p * 500 = 2500 Finally, we divide both sides by 500 to solve for p: p = 2500 / 500 p = 5 So, the certain percentage is boxed{5%} .