Example Question - approximation
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Division Approximation Problem
<p>Given the numbers 628941 and 128378, to find the approximation of the division of 628941 by 128378, we perform the following calculations:</p> <p>628941 \approx 629000</p> <p>128378 \approx 128000</p> <p>\frac{629000}{128000} = \frac{629}{128} \approx \frac{630}{130} = \frac{63}{13}</p> <p>63 \div 13 = 4 \text{ remainder } 11</p> <p>\frac{63}{13} \approx 4 + \frac{11}{13}</p> <p>Since \frac{11}{13} is close to \frac{12}{13}, which is approximately 1, we have:</p> <p>4 + \frac{11}{13} \approx 4 + 1 = 5</p> <p>Therefore, the approximation of the division is 5.</p>
Approximation to the Nearest Hundred
<p>Um eine Zahl auf den nächsten Hunderter zu runden, betrachten Sie die Zehnerstelle:</p> <p>Wenn die Zehnerstelle 5 oder höher ist, erhöhen Sie den Hunderter um eins und ändern alle folgenden Stellen zu 0.</p> <p>Wenn die Zehnerstelle 4 oder niedriger ist, ändern Sie alle folgenden Stellen zu 0 und lassen den Hunderter unverändert.</p> <p>628941 auf den nächsten Hunderter gerundet:</p> <p>Die Zehnerstelle ist "4", also lassen wir den Hunderter unverändert:</p> <p>628941 \approx 628900</p> <p>128378 auf den nächsten Hunderter gerundet:</p> <p>Die Zehnerstelle ist "7", also erhöhen wir den Hunderter um eins:</p> <p>128378 \approx 128400</p> <p>36951 auf den nächsten Hunderter gerundet:</p> <p>Die Zehnerstelle ist "5", also erhöhen wir den Hunderter um eins:</p> <p>36951 \approx 37000</p>
Approximating Values of Sine Using Right Triangle Ratios
<p>Para la parte a), usaremos la relación trigonométrica del seno, que es el lado opuesto sobre la hipotenusa en un triángulo rectángulo.</p> <p>\[\sen(\alpha) = \frac{opuesto}{hipotenusa}\]</p> <p>\[\sen(\alpha) = \frac{2\ mm}{4\ mm}\]</p> <p>\[\sen(\alpha) = 0.5\]</p> <p>Para la parte b), de igual manera, aplicamos la relación del seno:</p> <p>\[\sen(\alpha) = \frac{opuesto}{hipotenusa}\]</p> <p>\[\sen(\alpha) = \frac{4\ cm}{\sqrt{10}\ cm}\]</p> <p>\[\sen(\alpha) = \frac{4}{\sqrt{10}}\]</p> <p>\[\sen(\alpha) = \frac{4}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}\]</p> <p>\[\sen(\alpha) = \frac{4\sqrt{10}}{10}\]</p> <p>\[\sen(\alpha) = \frac{2\sqrt{10}}{5}\]</p> <p>Para obtener el valor aproximado a tres dígitos decimales, calculamos:</p> <p>\[\sen(\alpha) \approx \frac{2\sqrt{10}}{5} \approx 0.632\]</p>
Approximating the Area Under a Curve Using Riemann Sums
<p>لحل هذا السؤال، سنستخدم مجموع ريمان الأيسر:</p> <p>معادلة الدالة \( y = x^2 + 2 \)</p> <p>المجال هو \([0, 1]\)</p> <p>عدد المستطيلات \( n = 12 \)</p> <p>عرض كل مستطيل \( \Delta x = \frac{1 - 0}{12} = \frac{1}{12} \)</p> <p>نقاط التقييم اليسرى هي \( x_i = 0, \frac{1}{12}, \frac{2}{12}, \ldots, \frac{11}{12} \)</p> <p>الآن نقوم بحساب قيمة المجموع:</p> <p>\( \sum_{i=0}^{11} f(x_i) \Delta x \)</p> <p>\( = \sum_{i=0}^{11} \left( ( \frac{i}{12} )^2 + 2 \right) \frac{1}{12} \)</p> <p>\( = \frac{1}{12} \sum_{i=0}^{11} \left( \frac{i^2}{144} + 2 \right) \)</p> <p>\( = \frac{1}{12} \left( \sum_{i=0}^{11} \frac{i^2}{144} + \sum_{i=0}^{11} 2 \right) \)</p> <p>\( = \frac{1}{12} \left( \frac{1}{144} \sum_{i=0}^{11} i^2 + 2 \cdot 12 \right) \)</p> <p>حيث أن \( \sum_{i=0}^{11} i^2 \) هو مجموع مربعات العداد الأول 11 عدد طبيعي \( = 0^2 + 1^2 + 2^2 + \ldots + 11^2 \)</p> <p>\( = 0 + 1 + 4 + \ldots + 121 \)</p> <p>\( = 506 \)</p> <p>إذاً:</p> <p>\( \frac{1}{12} \left( \frac{506}{144} + 24 \right) \)</p> <p>\( = \frac{1}{12} \left( \frac{506}{144} + \frac{3456}{144} \right) \)</p> <p>\( = \frac{1}{12} \cdot \frac{3962}{144} \)</p> <p>\( = \frac{3962}{1728} \)</p> <p>\( \approx 2.292 \)</p> <p>النتيجة التقريبية لمساحة المنطقة تحت المنحنى هي \( \approx 2.292 \) وحدة مربعة.</p>
Euler Method for Solving Differential Equations
這是一個用 Euler 方法解微分方程的問題。給定的微分方程是： \[ \frac{dy}{dx} = 2xy - x \] 並且初始條件為 \( y(1) = 0 \)。問題是要利用 Euler 方法在 \( x = 1 \) 出發，並使用两个等大小的步伐來近似 \( y(0) \) 的值。 Euler 方法中，我們將從初始條件開始創建一个序列的估計值，每一步的形式是： \[ y_{i+1} = y_i + hf(x_i, y_i) \] 其中 h 是步長 (在這個例子中是 \( x \) 的改變量)，\( f(x, y) \) 是微分方程的右側 (在這個例子中是 \(2xy - x\))，而 \( y_i \) 和 \(x_i\) 是當前步的 \( y \) 和 \( x \) 的值。 步伐大小是 \( x \) 变化量的一半。因為我們從 \( x = 1 \) 出發並回到 \( x = 0 \)，因此每個步伐的大小应该是 \( h = \frac{0 - 1}{2} = -0.5 \)。 让我们开始计算： 第一步，從 \( x_0 = 1 \) 和 \( y_0 = 0 \) 開始: \[ y_1 = y_0 + h(2x_0y_0 - x_0) \] \[ y_1 = 0 + (-0.5)(2*1*0 - 1) \] \[ y_1 = 0 - 0.5(-1) \] \[ y_1 = 0.5 \] 這是在 \( x = 0.5 \) 的 \( y \) 的近似值。 第二步，使用 \( x_1 = 0.5 \) 和 \( y_1 = 0.5 \): \[ y_2 = y_1 + h(2x_1y_1 - x_1) \] \[ y_2 = 0.5 + (-0.5)(2*0.5*0.5 - 0.5) \] \[ y_2 = 0.5 - 0.5(0.5 - 0.5) \] \[ y_2 = 0.5 - 0.5(0) \] \[ y_2 = 0.5 \] 因此，根據 Euler 方法，\( y(0) \) 的近似值为\( y_2 = 0.5 \)，选项是 (B) \( \frac{1}{2} \)。
Calculating Cube Root of 16
To complete the statement, you need to calculate the cube root of 16 and then determine the numbers that are less than and greater than the obtained value. The cube root of 16, denoted as ∛16, is the number that when multiplied by itself three times gives the product 16. The exact cube root of 16 is not a whole number, but we can approximate it. We know that: ∛8 = 2 ∛27 = 3 Since 16 is between 8 and 27, the cube root of 16 must be between 2 and 3. Let's find an approximate value of ∛16. We know 2^3 is 8 and 2.5^3 (which is 2.5 * 2.5 * 2.5) equals 15.625. Since 15.625 is slightly less than 16, and increasing the base number to for example 2.6^3 will give us a number slightly more than 16 (approximately 17.576), we can conclude that ∛16 is slightly greater than 2.5 but less than 2.6. Filling in the blanks knowing ∛16 is greater than 2 and less than 3 would look something like this: 2 < ∛16 < 3 So, the statement filled out would be: 2 < ∛16 < 3
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