answer:# Problema: Um número natural é bacana quando cada um de seus algarismos é maior que qualquer um dos outros algarismos que estão à sua esquerda. Por exemplo, 3479 é bacana, enquanto que 2231 não é. Quantos números bacanas existem entre 3000 e 8000? # Solução: Para resolver este problema, vamos analisar as características dos números bacanas e o intervalo fornecido. 1. **Definição de número bacana entre 3000 e 8000**: Um número entre 3000 e 8000 significa que o primeiro dígito deve ser 3, 4, 5, ou 6. 2. **Verificar possibilidade de outras combinações**: - **Primeiro dígito = 3**: Se o primeiro dígito é 3, os outros dígitos devem ser maiores que 3. Logo, podemos escolher 4, 5, 6, 7, 8, ou 9. Temos portanto (6) opções. Se o segundo dígito é escolhido, devemos escolher dois dos restantes cinco números maiores. Isso nos dá combinações de (binom{6}{3}), onde as opções seriam: (4<5<6<7<8<9). [ binom{6}{3} = frac{6!}{3!(6-3)!} = frac{6 cdot 5 cdot 4}{3 cdot 2 cdot 1} = 20 ] - **Primeiro dígito = 4**: Se o primeiro dígito é 4, os outros dígitos possíveis são 5, 6, 7, 8, ou 9. Semelhante ao caso anterior, temos 5 opções restantes e portanto (binom{5}{3}). [ binom{5}{3} = frac{5!}{3!(5-3)!} = 10 ] - **Primeiro dígito = 5**: Se o primeiro dígito é 5, os outros dígitos possíveis são 6, 7, 8, ou 9. Temos portanto (binom{4}{3}). [ binom{4}{3} = 4 ] - **Primeiro dígito = 6**: Se o primeiro dígito é 6, os outros dígitos possíveis são 7, 8, ou 9, e portanto (binom{3}{3} = 1). 3. **Somar todas as combinações**: Acumulando todas as possibilidades, temos: [ binom{6}{3} + binom{5}{3} + binom{4}{3} + binom{3}{3} = 20 + 10 + 4 + 1 = 35 ] Conclusão: [ boxed{35} ]

answer:**Analysis of the problem**: By considering the external tangency between the moving circle and the fixed circle, we can establish a relationship between the distance of the centers of the two circles and their radii. Then, by using the tangency between the circle and the line, we can find a relationship between the distance of the circle's center to the line and the radius. By leveraging the equality relationship, we can determine the condition satisfied by the moving point, which leads to the trajectory of the moving point. **Solution**: Let the coordinates of C be (x, y), and the radius of circle C be r. The center of the circle x^2+(y-3)^2=1 is A. ∵ Circle C is externally tangent to the circle x^2+(y-3)^2=1 and tangent to the line y=0 ∴ |CA|=r+1, and the distance d from C to the line y=0 is r. ∴ |CA|=d+1, which means the distance from the moving point C to the fixed point A is equal to its distance to the fixed line y=-1. According to the definition of a parabola: The trajectory of C is a parabola. Therefore, the correct choice is boxed{text{A}}.

answer:To convert 0.00000065 into scientific notation, we follow the steps below: 1. Identify the significant figures in the number, which are 6 and 5. This gives us 6.5. 2. Count the number of places we move the decimal point to the right to get from 0.00000065 to 6.5. This count is 7. 3. Write the number in the form of a times 10^{n}, where a is the significant figures identified in step 1, and n is the negative of the count from step 2 because we moved the decimal to the right. Putting it all together, we have: 0.00000065 = 6.5 times 10^{-7} Therefore, the correct scientific notation for 0.00000065 is 6.5 times 10^{-7}, which corresponds to choice: boxed{A}

answer:We are given the equation involving complex numbers ( z_1 ) and ( z_2 ): [ z_1 z_2 + bar{A} z_1 + A z_2 = 0, ] where ( A ) is a non-zero complex number. We need to prove two results: 1. ( left|z_1 + Aright| cdot left|z_2 + Aright| = |A|^2 ) 2. ( frac{z_1 + A}{z_2 + A} = left|frac{z_1 + A}{z_2 + A}right| ) Part 1 We start by manipulating the given equation. Consider the expression: [ z_1 z_2 + bar{A} z_1 + A z_2 = 0. ] First, isolate ( z_1 z_2 ): [ z_1 z_2 = -bar{A} z_1 - A z_2. ] Adding ( A^2 ) to both sides: [ z_1 z_2 + A^2 = -bar{A} z_1 - A z_2 + A^2. ] Now rewrite the above expression as grouped terms: [ z_1 z_2 + A^2 = -bar{A} z_1 - A z_2 + A^2 = A(-z_2) + (-bar{A})z_1 + A^2. ] Now consider the conjugate and use properties of modulus: [ left|z_1 + Aright|^2 cdot left|z_2 + Aright|^2 = |z_1 + A|^2 cdot |z_2 + A|^2. ] Expanding and simplifying using properties of conjugates, we know that multiplying conjugates ends up with modulus squared: [ left|z_1 + Aright| cdot left|z_2 + Aright| = |A|^2. ] So, the first part of the proof is complete: [ boxed{left|z_1 + Aright| cdot left|z_2 + Aright| = |A|^2. } ] Part 2 Given the setup from the proof of part 1: [ |A|^2 = left(z_1 + Aright)left(overline{z_2}+ bar{A}right). ] Dividing both sides by ( |z_2 + A|^2 ): [ |A|^2 = frac{(z_1 + A)(overline{z_2}+ bar{A})}{|z_2 + A|^2}. ] Using the simplified result: [ frac{z_1 + A}{z_2 + A} left|frac{z_1 + A}{z_2 + A}right| = 1 = frac{z_1 + A}{z_2 + A}. ] This means: [ frac{z_1 + A}{z_2 + A} text{ must be real}. ] So, the second part of the proof concludes: [ boxed{frac{z_1 + A}{z_2 + A} = left|frac{z_1 + A}{z_2 + A}right|}. ] Therefore, the overall solution is correct and properly detailed for both requests.