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Example Question - mathematics

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Simplifying Mathematical Expressions

a) \left| \frac{2}{3} \cdot \left( -\frac{3}{5} \right) \cdot \frac{1}{4} \right| = \left| -\frac{2 \cdot 3}{3 \cdot 5 \cdot 4} \right| = \frac{6}{60} = \frac{1}{10} b) \left| \frac{1}{2} \cdot \left( \frac{2}{7} \right)^{\frac{3}{2}} \right| = \frac{1}{2} \cdot \frac{2^{\frac{3}{2}}}{7^{\frac{3}{2}}} = \frac{\sqrt{8}}{2\sqrt{343}} = \frac{2\sqrt{2}}{14\sqrt{7}} = \frac{\sqrt{2}}{7\sqrt{7}} c) \left| \left( \frac{5}{4} \right)^{-1} \right| = \left| \frac{4}{5} \right| = \frac{4}{5}

Calculating a Weighted Product

To solve the equation, first express the fraction: $\frac{9}{7} \times 32 \, \text{kg}$ Now, multiply the two values: $\frac{9 \times 32}{7} \, \text{kg}$ Calculating $9 \times 32 = 288$: $\frac{288}{7} \, \text{kg} \approx 41.14 \, \text{kg}$ Thus, the solution is approximately $41.14 \, \text{kg}$.

Calculating Exponential Expressions

Para calcular la expresión dada, primero resolvemos la parte exponencial: E = 10^{(32-25-8^{3-1})} + \left( \frac{1}{81} \right)^{(-16^{-4} \cdot 0.29,5)}. Realizamos las operaciones paso a paso: 1. Calculamos los exponentes y simplificamos: 32 - 25 = 7, 8^{3-1} = 8^{2} = 64. Entonces, E = 10^{(7-64)} + \left( \frac{1}{81} \right)^{(-16^{-4} \cdot 0.29,5)}. 2. Continuamos: 7 - 64 = -57, así que E = 10^{-57} + \left( \frac{1}{81} \right)^{(-16^{-4} \cdot 0.29,5)}. 3. Evaluamos \left( \frac{1}{81} \right)^{x}, donde x viene de la evaluación de -16^{-4} \cdot 0.29,5. 4. Sin embargo, con solo usar aproximaciones es difícil calcular sin más información. Finalmente, tras evaluaciones y estimaciones, las opciones dadas son: a) 12, b) 14, c) 11, d) 82, e) 13. Puede que la respuesta más cercana sea b) 14, depende del cálculo preciso.

Simplifying Algebraic Expressions

First, expand the numerator: (m^6 n)^2 = m^{12} n^2 Now the expression becomes: \frac{m^{12} n^2}{m^3 n^5} Next, apply the quotient rule: \frac{m^{12}}{m^3} = m^{12-3} = m^9 \frac{n^2}{n^5} = n^{2-5} = n^{-3} The expression simplifies to: m^9 n^{-3} In positive exponent form, it is: \frac{m^9}{n^3}

Simplifying a Fraction with Exponents

Given the expression: \[\frac{10^{-8}}{7^5 \times 10^3 \times 7^{-7}}\] We can rewrite the denominator: $7^5 \times 10^3 \times 7^{-7} = 10^3 \times 7^{5 - 7} = 10^3 \times 7^{-2}$ Next, we place this back into the expression: \[\frac{10^{-8}}{10^3 \times 7^{-2}} = \frac{10^{-8}}{10^3} \times 7^2\] Simplifying $\frac{10^{-8}}{10^3}$: $10^{-8 - 3} = 10^{-11}$ The expression becomes: $10^{-11} \times 7^2$ Thus, the final simplified expression is: \[7^2 \times 10^{-11}\] Which can be written as: $49 \times 10^{-11}$

Understanding Powers and Exponents

To simplify the expression, we recognize the pattern in the powers of -16. The expression can be rewritten as: (-16)1 × (-16)2 × (-16)3 × (-16)4 × (-16)5 This is equivalent to: (-16)1+2+3+4+5 Calculating the exponent: 1 + 2 + 3 + 4 + 5 = 15 So, the simplified expression is: (-16)15

Simplification of Exponential Expressions

To simplify $10^{-8}$, we can express it as: $10^{-8} = \frac{1}{10^{8}}$

Finding the Lowest Common Multiple

We need to find the LCM of 18, 36, and 54. Step 1: Factor each number into its prime factors: 18 = 2 \times 3^2 36 = 2^2 \times 3^2 54 = 2 \times 3^3 Step 2: Take the highest power of each prime: For 2: highest power is $2^2$ For 3: highest power is $3^3$ Step 3: Multiply these together: LCM = 2^2 \times 3^3 = 4 \times 27 = 108 The lowest common multiple (LCM) of 18, 36, and 54 is 108.

Finding the Inverse of a Function

1. Start with the function $ f(x) = \frac{x + 3}{x - 1} $. 2. Replace $ f(x) $ with $ y $: $ y = \frac{x + 3}{x - 1} $. 3. Swap $ x $ and $ y $: $ x = \frac{y + 3}{y - 1} $. 4. Multiply both sides by $ y - 1 $: $ x(y - 1) = y + 3 $. 5. Distribute: $ xy - x = y + 3 $. 6. Rearrange to solve for $ y $: $ xy - y = x + 3 $. 7. Factor out $ y $: $ y(x - 1) = x + 3 $. 8. Divide by $ x - 1 $: $ y = \frac{x + 3}{x - 1} $. 9. Therefore, the inverse function is $ f^{-1}(x) = \frac{x + 3}{x - 1} $.

Describing Real-Life Scenarios Using Algebraic Expressions

Dado que la imagen solicita describir una situación real para la expresión algebraica "2x", aquí hay un ejemplo de cómo podría ser esta situación: Si "x" representa la cantidad de horas trabajadas, entonces "2x" puede representar el salario ganado si una persona recibe un pago doble por cada hora trabajada.

Determine The Quantity of Digits for a Given Sequence Number

Untuk menyatakan bilangan angka bagi nombor berikut, kita perlu pertama menentukan apakah pola yang diikuti oleh nombor-nombor dalam siri tersebut. Walau bagaimanapun, soalan ini tidak menyediakan cukup maklumat untuk mengenal pasti siri nombor atau nombor tertentu yang diperlukan untuk ditentukan jumlah digitnya. Oleh itu, tidak mungkin untuk memberikan penyelesaian yang tepat tanpa maklumat tambahan. Tanpa informasi tambahan, kita tidak dapat menyatakan jumlah angka untuk nombor yang tidak diketahui.

Calculating the Cost of Bread and Gasoline Consumption

Para la pregunta 5, necesitamos hallar el costo si se hubieran comprado 45 barras en lugar de 70. Para ello, usaremos una regla de tres simple: \[ \text{Costo por 70 barras} : \text{número de barras} = \text{Costo por 45 barras} : 45 \] \[ 42\text{€} : 70 = x : 45 \] Multiplicamos en cruz: \[ 70x = 42 \cdot 45 \] \[ 70x = 1890 \] Ahora, despejamos x: \[ x = \frac{1890}{70} \] \[ x = 27 \text{€} \] Por lo tanto, si hubieran comprado 45 barras de pan, habrían pagado 27€. Para la pregunta 6, necesitamos calcular el coste del combustible en un viaje de 270km, manteniendo el consumo actual: \[ \frac{46 \text{céntimos}}{4\text{km}} = \frac{x \text{céntimos}}{270\text{km}} \] Multiplicamos en cruz para encontrar x: \[ 4x = 46 \cdot 270 \] \[ 4x = 12420 \] Ahora, despejamos x: \[ x = \frac{12420}{4} \] \[ x = 3105 \text{céntimos} \] Convertimos los céntimos a euros: \[ x = \frac{3105}{100} \] \[ x = 31.05\text{€} \] Por lo tanto, el costo del combustible para un viaje de 270km sería 31.05€.

Solving Word Problems Involving Proportions and Unit Costs

Para la primera pregunta: Si 70 barras de pan cuestan 42€, entonces el costo por barra es $ \frac{42€}{70} $ Para saber cuánto costarían 45 barras: $ 45 \times \frac{42€}{70} $ $ = 45 \times 0.6€ $ $ = 27€ $ Para la segunda pregunta: Si gastar 46 céntimos de euro de gasolina para recorrer 4km, entonces el costo por kilómetro es $ \frac{46}{4} $ céntimos. Para saber cuánto costaría el combustible en un viaje de 270km: $ 270 \times \frac{46}{4} $ $ = 270 \times 11.5 $ $ = 3105 $ céntimos, lo que equivale a $ 31.05€ $ Para la tercera pregunta: Si el precio de 15 menús es 120€, entonces el costo por menú es $ \frac{120€}{15} $ Para 7 personas: $ 7 \times \frac{120€}{15} $ $ = 7 \times 8€ $ $ = 56€ $

Determining the Number of Teachers Teaching Physics

Let $P$ be the number of teachers who teach physics and $M$ be the number of teachers who teach mathematics. We are given that there are 20 teachers in total, of which 12 teach mathematics, 4 teach both physics and mathematics. We need to find how many teach physics. We can use the principle of inclusion and exclusion to find the number of teachers who teach only physics. Number of teachers teaching only mathematics is $M - 4$. Number of teachers teaching physics, including those who teach both subjects, is $P$. The sum of teachers teaching only mathematics, only physics, and both is the total number of teachers: $(M - 4) + P = 20$ Given that $M = 12$, we substitute this into the equation: $(12 - 4) + P = 20$ $8 + P = 20$ $P = 20 - 8$ $P = 12$ There are 12 teachers who teach physics.

Place Value Exchange in Numbers

Soruyu çözmek için her bir sayıdaki basamak değerlerini bir yer sağa veya sola taşıyacağız. a) 2T 36H 1Z 24E için: 2T (yani, 2 x 10^3) 20H'ye dönüşür. 36H (yani, 36 x 10^2) 360Z'e dönüşür. 1Z (yani, 1 x 10^1) 10E'ye dönüşür. 24E (yani, 24 x 10^0) aynı kalır çünkü bunu taşırsak sayının değeri değişir. Bu durumda, yeni sayı: 20H + 360Z + 10E + 24E = 20H 360Z 34E olarak yazılabilir. b) 8ZT 0T 4H 22Z 1E için: 8ZT (yani, 8 x 10^4) 80T'ye dönüşür. 0T değişmez çünkü zaten 0 değerindedir. 4H (yani, 4 x 10^2) 40Z'ye dönüşür. 22Z (yani, 22 x 10^1) 220E'ye dönüşür. 1E (yani, 1 x 10^0) aynı kalır. Yeni sayı: 80T + 0T + 40Z + 220E + 1E = 80T 40Z 221E olarak yazılabilir. c) 3ZT 56T 2H 57E için: 3ZT (yani, 3 x 10^5) 30T'ye dönüşür. 56T (yani, 56 x 10^3) 560H'ye dönüşür. 2H (yani, 2 x 10^2) 20Z'ye dönüşür. 57E (yani, 57 x 10^0) aynı kalır. Yeni sayı: 30T + 560H + 20Z + 57E = 30T 560H 20Z 57E olarak yazılabilir.

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Example Question - mathematics

Here are examples of questions we've helped users solve.

Simplifying Mathematical Expressions

Calculating a Weighted Product

Calculating Exponential Expressions

Simplifying Algebraic Expressions

Simplifying a Fraction with Exponents

Understanding Powers and Exponents

Simplification of Exponential Expressions

Finding the Lowest Common Multiple

Finding the Inverse of a Function

Describing Real-Life Scenarios Using Algebraic Expressions

Determine The Quantity of Digits for a Given Sequence Number

Calculating the Cost of Bread and Gasoline Consumption

Solving Word Problems Involving Proportions and Unit Costs

Determining the Number of Teachers Teaching Physics

Place Value Exchange in Numbers

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