MIT

🔍

question:Given the coordinates of (A), (B), and (C) are respectively (A(4,0)), (B(0,4)), and (C(3cos alpha,3sin alpha)). ((1)) If (alpha in (pi,0)) and (|overrightarrow{AC}|=|overrightarrow{BC}|), find the value of angle (alpha); ((2)) If (overrightarrow{AC} perp overrightarrow{BC}), find the value of (frac{2sin^2alpha+sin 2alpha}{1+tan alpha}).

answer:Solution: ((1)) (overrightarrow{AC}=(3cos alpha-4,3sin alpha)), (overrightarrow{BC}=(3cos alpha,3sin alpha-4)) (overrightarrow{AC}^2=25-24cos alpha), (overrightarrow{BC}^2=25-24sin alpha) Since (|overrightarrow{AC}|=|overrightarrow{BC}|), (25-24cos alpha=25-24sin alpha) (sin alpha=cos alpha) Given (alpha in (-pi,0)), (alpha= -frac{3}{4}pi). So, the answer is boxed{alpha= -frac{3}{4}pi}. ((2)) Since (overrightarrow{AC} perp overrightarrow{BC}), (overrightarrow{AC} cdot overrightarrow{BC}=0) ((3cos alpha-4) times 3cos alpha+3sin alpha times (3sin alpha-4)=0) Solving gives (sin alpha+cos alpha= frac{3}{4}) Thus, (1+2sin alphacos alpha= frac{9}{16}) (2sin alphacos alpha= -frac{7}{16}) Therefore, (frac{2sin^2alpha+sin 2alpha}{1+tan alpha} = frac{2sin alpha(sin alpha+cos alpha)}{frac{sin alpha+cos alpha}{cos alpha}}=2sin alphacos alpha= -frac{7}{16}). So, the answer is boxed{-frac{7}{16}}.

question:Define F(a, b, c) = a times b^{c+1}. What positive value of s is the solution to the equation F(s, s, 2) = 1296?

answer:The function F(a, b, c) = a times b^{c+1} applied to s gives: [ F(s, s, 2) = s times s^{2+1} = s times s^3 = s^4. ] So, we have the equation: [ s^4 = 1296. ] To solve for s, we recognize that 1296 = 6^4, so: [ s = 6. ] Thus, the solution to the equation is s = boxed{6}.

question:A pentagon is drawn on a square grid. Determine how many of the five triangles formed by connecting any three vertices of this pentagon are isosceles. Vertices of the pentagon are given as follows: (A(0, 0)) (B(2, 4)) (C(5, 4)) (D(7, 0)) (E(3, 1)) A) 0 B) 1 C) 2 D) 3 E) 4

answer:To determine how many of the triangles are isosceles, we calculate the lengths of the sides of each triangle formed by any three of the vertices (A, B, C, D,) and (E) and check if at least two sides are equal in length. Let's calculate the distances between vertices: - (AB = sqrt{(2-0)^2 + (4-0)^2} = sqrt{4+16} = sqrt{20}) - (AC = sqrt{(5-0)^2 + (4-0)^2} = sqrt{25+16} = sqrt{41}) - (AD = sqrt{(7-0)^2 + (0-0)^2} = sqrt{49} = 7) - (AE = sqrt{(3-0)^2 + (1-0)^2} = sqrt{9+1} = sqrt{10}) - (BC = sqrt{(5-2)^2 + (4-4)^2} = sqrt{9} = 3) - (BD = sqrt{(7-2)^2 + (0-4)^2} = sqrt{25+16} = sqrt{41}) - (BE = sqrt{(3-2)^2 + (1-4)^2} = sqrt{1+9} = sqrt{10}) - (CD = sqrt{(7-5)^2 + (0-4)^2} = sqrt{4+16} = sqrt{20}) - (CE = sqrt{(3-5)^2 + (1-4)^2} = sqrt{4+9} = sqrt{13}) - (DE = sqrt{(3-7)^2 + (1-0)^2} = sqrt{16+1} = sqrt{17}) Now we form and check each triangle: 1. **Triangle ABC**: - (AB = sqrt{20}), (AC = sqrt{41}), (BC = 3) - Not isosceles. 2. **Triangle ABD**: - (AB = sqrt{20}), (AD = 7), (BD = sqrt{41}) - Not isosceles. 3. **Triangle ABE**: - (AB = sqrt{20}), (AE = sqrt{10}), (BE = sqrt{10}) - Isosceles (AE = BE). 4. **Triangle BCD**: - (BC = 3), (BD = sqrt{41}), (CD = sqrt{20}) - Not isosceles. 5. **Triangle CDE**: - (CD = sqrt{20}), (CE = sqrt{13}), (DE = sqrt{17}) - Not isosceles. Conclusion: Only Triangle ABE is isosceles. Thus, the number of isosceles triangles is 1. The final answer is boxed{B}

question:Find the remainder when (1^3 + 2^3 + 3^3 + dots + 150^3) is divided by 9.

answer:1. **Compute the cubes modulo 9**: Evaluating the first few cubes (n^3) modulo 9: begin{align*} 1^3 & equiv 1 pmod{9}, 2^3 & equiv 8 pmod{9}, 3^3 & equiv 27 equiv 0 pmod{9}, 4^3 & equiv 64 equiv 1 pmod{9}, 5^3 & equiv 125 equiv 8 pmod{9}, 6^3 & equiv 216 equiv 0 pmod{9}, 7^3 & equiv 343 equiv 1 pmod{9}, 8^3 & equiv 512 equiv 8 pmod{9}, 9^3 & equiv 729 equiv 0 pmod{9}. end{align*} From the calculation, notice that (n^3 equiv n pmod{9}) for (n = 1, 2, ldots, 9) and those multiples of 9. 2. **Sum the sequence up to 150**: - The sequence (1^3, 2^3, ldots, 9^3) repeats every 9 terms. Hence, we consider the sum (1+2+0+1+8+0+1+8+0 = 21) for each group of 9 numbers. - Count of such full groups in 150 numbers: (frac{150}{9} = 16.) Full sets sum: (21 times 16 = 336.) - Summing the remaining terms (from 136 to 150): Compute directly for numbers equivalent to 1 to 15 mod 9 (1,2,4,5,7,8): (1^3 + 2^3 + 1^3 + 8^3 + 1^3 + 8^3 pmod{9}equiv 1 + 8 + 1 + 8 + 1 + 8 = 27 equiv 0 pmod{9}). - Total mod 9: (336 + 0 equiv 6 pmod{9}). Conclusion: The remainder when (1^3 + 2^3 + 3^3 + dots + 150^3) is divided by 9 is (boxed{6}).