February 2026

• Feb 24 $ 2|y| \leq 8 \Rightarrow |y| \leq 4 $
• Feb 24 $ 2|x| \leq 8 \Rightarrow |x| \leq 4 $
• Feb 24 $ y = 3 \Rightarrow x = 9 $
• Feb 24 $ y = 2 \Rightarrow x = 4 $
• Feb 24 $ y = 0 \Rightarrow x = 0 $
• Feb 24 = 3r^2
• Feb 24 🌟 Yukon Gold Potato Recipes That Will Transform Your Dinner Game Overnight!
• Feb 24 😲 Yuki Sohma Shocked the World—Here’s What She Revealed!
• Feb 24 #### 1000An archaeologist discovers a site with 300 pottery shards. If 15% are from a specific culture, and 20% of those are intact, how many intact shards are from that culture?
• Feb 24 #### 10000000
• Feb 24 #### 11460
• Feb 24 #### 74.08
• Feb 24 #### 61.90
• Feb 24 #### ### This YSL Clutch Is a Game-Changer – Luxe, Sleek, and Unforgettable!
• Feb 24 الإجابة النهائية:** $\boxed{30^\circ}$
• Feb 24 الأجزاء التخيلية هي $-\frac{1}{2}$ و $-\frac{1}{2}$ على التوالي.لذلك، الجزء التخيلي الأكبر هو $\frac{1}{2} = \sin 30^\circ$.
• Feb 24 الأجزاء التخيلية هي $\sin \frac{\pi}{6} = \frac{1}{2}$ و $\sin \frac{5\pi}{6} = \frac{1}{2}$. للجذور من $z^2 = \frac{1 - i\sqrt{3}}{2} = e^{-i\pi/3}$, فإن الجذور التربيعية هي:
• Feb 24 إذن، $z^2 = \frac{1 \pm i\sqrt{3}}{2}$. الجذور التربيعية لـ $\frac{1 + i\sqrt{3}}{2}$ (التي تُعتبر $e^{i\pi/3}$) هي:
• Feb 24 لنفترض $w = z^2$, إذن:
• Feb 24 السؤال: من بين جميع جذور المعادلة $z^4 - z^2 + 1 = 0$, يمكن التعبير عن الجزء التخيلي الأكبر من جذر على شكل $\sin \theta$, حيث $\theta \in [0^\circ, 90^\circ]$. أوجد $\theta$.
• Feb 24 الإجابة النهائية:** الزوايا هي $\boxed{30^\circ}$, $\boxed{150^\circ}$, و $\boxed{270^\circ}$.
• Feb 24 وبالتالي، الحلول في $[0^\circ, 360^\circ]$ هي:
• Feb 24 لـ $\sin z = -1$:
• Feb 24 لـ $\sin z = \frac{1}{2}$:
• Feb 24 إذن، $u = \frac{1}{2}$ أو $u = -1$.
• Feb 24 لنفترض $u = \sin z$, إذن:
• Feb 24 - 2\sin^2 z = \sin z \implies 2\sin^2 z + \sin z - 1 = 0
• Feb 24 باستخدام متطابقة الزاوية المزدوجة، $\cos 2z = 1 - 2\sin^2 z$, تصبح المعادلة:
• Feb 24 السؤال: أوجد جميع الزوايا $z \in [0^\circ, 360^\circ]$ التي تحقق المعادلة $\cos 2z = \sin z$.
• Feb 24 الإجابة النهائية:** $\boxed{\begin{pmatrix} t \\ 2 \\ 3 \end{pmatrix}}$ لأي عدد حقيقي $t$.
• Feb 24 إذن، $\mathbf{v} = \begin{pmatrix} v_1 \\ 2 \\ 3 \end{pmatrix}$، حيث يمكن أن يكون $v_1$ أي عدد حقيقي لأنه لا يؤثر على الضرب التقاطعي مع $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$. للحصول على حل خاص، نضع $v_1 = 0$:
• Feb 24 -v_2 = -2 \implies v_2 = 2
• Feb 24 نحتاج إلى أنه يساوي $\begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix}$, مما يعطي:
• Feb 24 الحل: ليكن $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$. حاصل الضرب التقاطعي هو:
• Feb 24 السؤال: أوجد المتجه $\mathbf{v}$ بحيث أن $\mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix}$.
• Feb 24 الإجابة النهائية:** الزوايا هي $\boxed{240^\circ}$ و $\boxed{300^\circ}$.
• Feb 24 الزوايا $z$ التي تحقق $\sin z = -\frac{\sqrt{3}}{2}$ تقع في الربعين الثالث والرابع. الزاوية المرجعية هي $60^\circ$ لأن $\sin 60^\circ = \frac{\sqrt{3}}{2}$. لذلك، الحلول في الفترة $[0^\circ, 360^\circ]$ هي:
• Feb 24 السؤال: أوجد جميع الزوايا $z \in [0^\circ, 360^\circ]$ التي تحقق المعادلة $2\sin z + \sqrt{3} = 0$.
• Feb 24 الآن، لنفترض $c = 3$, إذن:
• Feb 24 الحل: ليكن $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. نحتاج إلى إيجاد $a$, $b$, $c$, و $d$ بحيث:
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• Feb 24 -v_1 - 2v_2 = -2
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• Feb 24 – Yasmine Malek Reveals the Dark Secrets Hidden Behind Her Fame!
• Feb 24 #### 1285.71
• Feb 24 \( |17,300 - 15,000| = 2,300 \)
• Feb 24 \( |15,900 - 15,000| = 900 \)
• Feb 24 \( |16,400 - 15,000| = 1,400 \)
• Feb 24 \( |15,100 - 15,000| = 100 \)
• Feb 24 \( |13,800 - 15,000| = 1,200 \)
• Feb 24 \( |14,200 - 15,000| = 800 \)
• Feb 24 \( |12,500 - 15,000| = 2,500 \)
• Feb 24 \( z = \frac{20 - 14}{3} = 2 \)
• Feb 24 \( (76-74)^2 = 4 \)
• Feb 24 \( (82-74)^2 = 64 \)
• Feb 24 \( (70-74)^2 = 16 \)
• Feb 24 \( (74-74)^2 = 0 \)
• Feb 24 \( (68-74)^2 = 36 \)
• Feb 24 #### 48720
• Feb 24 \( P(20) \approx 1200 \times 40.6 = 48,720 \)
• Feb 24 \( e^{3.712} \approx e^{3.7} \approx 40.45 \) (Since \( e^{3.4} \approx 30, e^{3.7} \approx 40.5, e^{3.712} \approx 40.6 \))
• Feb 24 \( P(20) = 1200 e^{0.1856 \times 20} = 1200 e^{3.712} \)
• Feb 24 \( 10k = \ln(6.4) \approx 1.856 \Rightarrow k \approx 0.1856 \)
• Feb 24 \( 7680 = 1200 e^{10k} \Rightarrow e^{10k} = \frac{7680}{1200} = 6.4 \)
• Feb 24 \( P_0 = 1200 \), \( P(10) = 7680 \)
• Feb 24 #### 930 and 460
• Feb 24 \( \frac{t}{3} \geq 13.9657 \Rightarrow t \geq 41.897 \)
• Feb 24 \( \log_2(10^3) = 3 \log_2(10) \approx 3 \times 3.3219 = 9.9657 \)
• Feb 24 \( \log_2(16,000) = \log_2(16 \times 1000) = \log_2(16) + \log_2(1000) = 4 + \log_2(10^3) \)
• Feb 24 #### 58.33
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• Feb 24 \( b_3 = M(b_2) = M\left(\frac{3}{4}\right) = \frac{3}{4} - \frac{\left(\frac{3}{4}\right)^4}{4} \)
• Feb 24 \( b_2 = M(b_1) = M(1) = 1 - \frac{1^4}{4} = 1 - \frac{1}{4} = \frac{3}{4} \)
• Feb 24 = \left( 8 - \frac{16}{3} \right) = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}.
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• Feb 24 = 2^4 \times 3,\quad 72 = 2^3 \times 3^2
• Feb 24 #### $9070.58Question: What is the smallest positive integer that is divisible by both the number of continents and the number of oceans on Earth, assuming a total of 7 continents and 5 oceans?
• Feb 24 #### Width: 56.315 m, Length: 106.315 m
• Feb 24 #### 2.4 hours
• Feb 24 #### 28.28 cm
• Feb 24 #### 144 units
• Feb 24 #### 153.86 cm²
• Feb 24 #### Width: 12 meters, Length: 24 meters
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• Feb 24 ÷ 1.25 = 680 ÷ 100 = 6.8 — correct.
• Feb 24 \( \frac{\text{processing time}}{\text{interarrival time}} = \frac{8.5}{1.25} = 6.8 \)
• Feb 24 \( \frac{8.5}{1.25} = 6.8 \). However, in real systems, it’s approximate. But math olympiad may accept decimal.
• Feb 24 \( 8.5 \div 1.25 = 6.8 \). But let’s re-express:
• Feb 24 \( 8.5 / 1.25 = 6.8 \), but since parts are discrete, use exact value. However, previous examples use exact math. So:
• Feb 24 #### 92.31
• Feb 24 – X Games 2025 Shakes Up Gaming with Epic New Titles & Unleashed Action!
• Feb 24 \( u = 3 \Rightarrow v = 9 \)
• Feb 24 \( u = 2 \Rightarrow v = 4 \)
• Feb 24 \( u = 0 \Rightarrow v = 0 \)
• Feb 24 \( f(2) = -4 -4 + 4 = -4 \) ✓
• Feb 24 \( f(1) = -1 -2 + 4 = 1 \) ✓
• Feb 24 \( f(0) = 4 \) ✓
• Feb 24 \( f(2) = 4a + 2b + c = -4 \) → \( 4a + 2b + 4 = -4 \) → \( 4a + 2b = -8 \) → \( 2a + b = -4 \)
• Feb 24 \( f(1) = a + b + c = 1 \) → \( a + b + 4 = 1 \) → \( a + b = -3 \)
• Feb 24 \( f(0) = c = 4 \)
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• Feb 24 $ \sin(4\pi/9) = \sin(80^\circ) \approx 0.9848 $
• Feb 24 $ \sin(-8\pi/9) = -\sin(40^\circ) $
• Feb 24 $ \sin(-2\pi/9) = -\sin(40^\circ) $
• Feb 24 $ \sin(14\pi/9) = \sin(280^\circ) = \sin(-80^\circ) = -\sin(80^\circ) $ (negative)
• Feb 24 $ \sin(8\pi/9) = \sin(160^\circ) = \sin(20^\circ) \approx 0.3420 $
• Feb 24 $ \sin(2\pi/9) = \sin(40^\circ) \approx 0.6428 $
• Feb 24 - \frac{3}{2}(1) = -\frac{1}{2}.
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• Feb 24 > 7000, so on day 28.
• Feb 24 #### 816
• Feb 24 #### 3002
• Feb 24 #### 480A wildlife conservation team tracked 120 endangered turtles in a coastal reserve. After one year, the population increased by 25%, but then 30 turtles were relocated to a safe habitat center. The following year, the remaining population grew by 20%. What is the turtle population at the end of the second year?
• Feb 24 #### 493.375
• Feb 24 – 250 × 1.9735 ≈ 493.375 MPa
• Feb 24 θ = arctan(0.2) ≈ 11.3°
• Feb 24 = (4.8 / 3.2) × 10⁵ = 1.5 × 10⁵ = 150,000
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• Feb 24 \( xD(N^2 + D^2) = x(x^2 + 1)(x^6 - 5x^4 + 11x^2 + 1) \)
• Feb 24 \( N^2 + D^2 = x^6 - 6x^4 + 9x^2 + x^4 + 2x^2 + 1 = x^6 - 5x^4 + 11x^2 + 1 \)
• Feb 24 \( D^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1 \)
• Feb 24 \( N^3 = (x^3 - 3x)^3 = x^9 - 9x^7 + 27x^5 - 27x^3 \)
• Feb 24 \( N^2 = (x^3 - 3x)^2 = x^6 - 6x^4 + 9x^2 \)
• Feb 24 \( bc + a^2 = 1 \)
• Feb 24 \( ac + cd = c(a + d) = 0 \)
• Feb 24 \( ab + bd = b(a + d) = 0 \)
• Feb 24 \( bc + d^2 = 1 \)
• Feb 24 \( ac + cd = 0 \)
• Feb 24 \( ab + bd = 0 \)
• Feb 24 \( a^2 + bc = 1 \)
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