answer:If you incorrectly subtracted 152 to get 346, the number you started with before the incorrect subtraction was: 346 + 152 = 498 Now, to find the correct number, you should add 152 instead of subtracting it: 498 + 152 = 650 So the correct number would be boxed{650} .

answer:To determine the sum of all real numbers x satisfying the equation [(x^2-4x+2)^{x^2-5x+2} = 1,] we consider the conditions under which a^b = 1 for real numbers a and b. These conditions are either the exponent b=0, or the base a=1 or a=-1. Additionally, if a=-1, then b must be an even integer. # Vieta's Formulas For a quadratic equation x^2 + bx + c = 0, the sum of its roots, denoted as p+q, is -b, and the product of its roots, denoted as pq, is c. This is derived from the fact that (x-p)(x-q) = x^2 - (p+q)x + pq. If the leading coefficient is not 1, for an equation ax^2 + bx + c = 0, dividing the entire equation by a gives x^2 + frac{b}{a}x + frac{c}{a} = 0, where the sum of roots is -frac{b}{a} and the product is frac{c}{a}. # Case Analysis **First Case:** The exponent is 0 when x^2-5x+2=0. The discriminant is 5^2-4cdot1cdot2=17, indicating two distinct real roots. By Vieta's formulas, the sum of these roots is 5. These roots do not satisfy x^2-4x+2=0, avoiding the indeterminate form 0^0. Thus, the sum of solutions in this case is 5. **Second Case:** The base is 1 when x^2-4x+2=1, simplifying to x^2-4x+1=0. This equation also has a positive discriminant, implying two real roots. By Vieta's formulas, the sum of these roots is 4. Since 1^b=1 for any real b, both roots are solutions to the original equation. The sum of solutions in this case is 4. **Third Case:** The base is -1 when x^2-4x+2=-1, simplifying to x^2-4x+3=0. This factors into (x-1)(x-3)=0, yielding roots x=1 and x=3. Checking these values, we find that (-1)^{-2}=1 and (-1)^{-4}=1, confirming both as solutions. The sum of solutions in this case is 1+3=4. Adding the sums from all cases, we get 5 + 4 + 4 = 13. Therefore, the sum of all real numbers x satisfying the given equation is boxed{13}.

answer:Let's denote the initial number of dogs as D. After one month, 50% of the dogs are adopted, so there are 0.5D dogs left. For the cats, 25% of the 28 cats are adopted, so there are 0.75 * 28 cats left. That's 21 cats. For the lizards, 20% of the 20 lizards are adopted, so there are 0.8 * 20 lizards left. That's 16 lizards. The shelter takes in 13 new pets a month, so we add 13 to the remaining animals. The total number of pets in the shelter after one month is the sum of the remaining dogs, cats, lizards, and the new pets: 0.5D + 21 + 16 + 13 = 65 Now, let's solve for D: 0.5D + 50 = 65 0.5D = 65 - 50 0.5D = 15 D = 15 / 0.5 D = 30 Initially, there were boxed{30} dogs in the shelter.

answer:1. **Understanding the Problem:** We are given a triangle ( triangle ABC ) with the incenter ( O ), and the extensions of ( AO, BO, ) and ( CO ) intersect the circumcircle of ( triangle ABC ) at points ( A_1, B_1, ) and ( C_1 ) respectively. We need to prove that the area of ( triangle A_1B_1C_1 ) is T = frac{R^2}{2} (sin alpha + sin beta + sin gamma), where ( R ) is the radius of the circumcircle and ( alpha, beta, ) and ( gamma ) are the angles of ( triangle ABC ). 2. **Known Formula for Triangle Area:** It is known that the area of any triangle inscribed in a circle with radius ( R ) is given by T = frac{R^2}{2} (sin 2alpha + sin 2beta + sin 2gamma), where ( alpha, beta, gamma ) are the internal angles of the triangle. 3. **Establish Relations Between Angles:** To find the internal angles of ( triangle A_1B_1C_1 ), we analyze the angles formed by the intersections of circles. Since ( A_1, B_1, C_1 ) are intersections of the extensions of the medians with the circumcircle, we know: [ alpha_1 = frac{beta + gamma}{2}, quad beta_1 = frac{alpha + gamma}{2}, quad gamma_1 = frac{alpha + beta}{2}, ] where ( alpha_1, beta_1, gamma_1 ) are the internal angles of ( triangle A_1B_1C_1 ). 4. **Express the Triangle Area Formula:** Using the formula for the area of a triangle inscribed in a circle, the area ( T ) of ( triangle A_1B_1C_1 ) can be expressed as: T = frac{R^2}{2} (sin 2 alpha_1 + sin 2 beta_1 + sin 2 gamma_1). 5. **Simplify Using Trigonometric Identities:** Substitute the expressions for ( alpha_1, beta_1, gamma_1 ) into the formula and use the angle addition formula for sine: [ sin(2alpha_1) = sin(beta + gamma), quad sin(2beta_1) = sin(alpha + gamma), quad sin(2gamma_1) = sin(alpha + beta). ] 6. **Use Identity for Sine of Supplementary Angles:** Remember that ( sin(180^circ - x) = sin x ): [ sin(2alpha_1) = sin(beta + gamma) = sin(180^circ - alpha), quad sin(2beta_1) = sin(alpha + gamma) = sin(180^circ - beta), quad sin(2gamma_1) = sin(alpha + beta) = sin(180^circ - gamma). ] 7. **Substitute and Simplify:** Substituting these back into the area formula gives: [ begin{aligned} T & = frac{R^2}{2} (sin 2alpha_1 + sin 2beta_1 + sin 2gamma_1) & = frac{R^2}{2} (sin(beta + gamma) + sin(alpha + gamma) + sin(alpha + beta)) & = frac{R^2}{2} (sin(180^circ - alpha) + sin(180^circ - beta) + sin(180^circ - gamma)) & = frac{R^2}{2} (sin alpha + sin beta + sin gamma). end{aligned} ] 8. **Conclusion:** Therefore, the area of ( triangle A_1B_1C_1 ) is given by boxed{frac{R^2}{2} (sin alpha + sin beta + sin gamma)}.