<p>The given expression is \( \sqrt[3]{-9} \times \sqrt[3]{\sqrt[3]{3}} \).</p> <p>Let's simplify the expression step by step.</p> <p>\( \sqrt[3]{-9} \) can be written as \( -\sqrt[3]{9} \) because \( \sqrt[3]{-1} \) equals -1.</p> <p>Now, \( -\sqrt[3]{9} \times \sqrt[3]{\sqrt[3]{3}} \) can be simplified further.</p> <p>Since \( \sqrt[3]{9} = \sqrt[3]{3^2} \), we can replace \( \sqrt[3]{9} \) with \( 3^{2/3} \), giving us:</p> <p>\( -3^{2/3} \times \sqrt[3]{\sqrt[3]{3}} \)</p> <p>We now focus on \( \sqrt[3]{\sqrt[3]{3}} \), which simplifies to \( \sqrt[3]{3^{1/3}} \).</p> <p>Using the property \( \sqrt[n]{a^m} = a^{m/n} \), we get:</p> <p>\( \sqrt[3]{3^{1/3}} = 3^{(1/3)(1/3)} = 3^{1/9} \).</p> <p>So the expression becomes:</p> <p>\( -3^{2/3} \times 3^{1/9} \)</p> <p>Now, we combine the exponents of the same base (3) using the law of exponents \( a^m \times a^n = a^{m+n} \):</p> <p>\( -3^{2/3 + 1/9} \)</p> <p>We find a common denominator for the exponents:</p> <p>\( -3^{6/9 + 1/9} \)</p> <p>\( -3^{7/9} \)</p> <p>Therefore, the simplified expression is \( -3^{7/9} \).</p>
\[ \mathrm{Given:} \sqrt[3]{x^9y^3} \quad \mathrm{and} \quad x, y \text{ are positive} \] \[ \sqrt[3]{x^9y^3} = \sqrt[3]{(x^3)^3y^3} = x^3 \cdot \sqrt[3]{y^3} = x^3y \] \[ \mathrm{Answer:} \quad x^3y \]
The image contains a mathematical question asking for another name for the expression \( B^{3/9} \). Simplifying \( B^{3/9} \) by dividing both the numerator and the denominator of the exponent by their greatest common divisor, which is 3, we get \( B^{(3÷3)/(9÷3)} = B^{1/3} \). Now let's look at the answer options: A) \( B^2 \) B) \( \sqrt[3]{B} \) C) \( B^9 \) D) \( \sqrt[9]{B} \) The correct answer is B, \( \sqrt[3]{B} \), because an exponent of \( 1/3 \) is equivalent to the cube root of the base, in this case, \( B \).
Claro, vamos a resolver la expresión paso a paso. Primero, abordemos la parte izquierda de la ecuación: \[ \sqrt[3]{\frac{3^4}{2^2}} \cdot \sqrt[3]{\sqrt[3]{64}} = \sqrt[3]{\frac{81}{4}} \cdot \sqrt[3]{\sqrt[3]{4^3}} \] Como \(3^4 = 81\) y \(2^2 = 4\), la primera raíz cúbica se simplifica a \(\sqrt[3]{\frac{81}{4}}\). Además, como \(64 = 4^3\), la segunda expresión se simplifica a \(\sqrt[3]{4}\): \[ \sqrt[3]{\frac{81}{4}} \cdot \sqrt[3]{4} = \frac{\sqrt[3]{81}}{\sqrt[3]{4}} \cdot \sqrt[3]{4} = \sqrt[3]{81} = 3^{4/3} = 3^{1\frac{1}{3}} = 3 \cdot 3^{1/3} \] Ahora, abordemos la parte derecha de la ecuación: \[ \sqrt{2^2 + 6} = \sqrt{4 + 6} = \sqrt{10} \] Ahora, sumamos las dos partes para obtener la expresión completa: \[ 3 \cdot 3^{1/3} + \sqrt{10} \] Y esta es la forma simplificada de la expresión original. No se puede simplificar más sin conocer el valor decimal de las raíces.
The expression presented in the image is the cube root of 976, denoted as \( \sqrt[3]{976} \). To find which integer is closest to this value, you would generally have to evaluate this cube root. You know that: \(2^3 = 8\), \(3^3 = 27\), \(4^3 = 64\), \(5^3 = 125\), \(6^3 = 216\), \(7^3 = 343\), \(8^3 = 512\), \(9^3 = 729\), \(10^3 = 1000\). Since 976 is between \(9^3 = 729\) and \(10^3 = 1000\), the cube root of 976 will be between 9 and 10. Because 976 is closer to \(9^3 = 729\) than to \(10^3 = 1000\), the integer closest to \( \sqrt[3]{976} \) is 9.
To solve the question: "Which integer is closest to cubed root of 976?" we will look for a perfect cube close to 976. We may not be able to find the exact cubed root instantly, but we can approximate by considering the cubes of integers. For example: - \( 9^3 = 729 \) - \( 10^3 = 1000 \) Given that 976 lies between 729 and 1000, we can determine that the cube root of 976 is between 9 and 10. But because 976 is closer to 1000 than it is to 729, it's cube root will be closer to 10. Therefore, the integer closest to the cube root of 976 is 10.
The image is asking for the integer closest to the cube root of 111. To solve this, we should look for two perfect cubes that are close to 111 and then find their cube roots. We know that: - \(4^3 = 64\) - \(5^3 = 125\) Since 111 is between 64 and 125, the cube root of 111 must be between 4 and 5. As 111 is closer to 125 than to 64, the cube root of 111 is closer to 5 than to 4. Hence, the closest integer to \(\sqrt[3]{111}\) is 5.
The expression in the question is the cube root of √2, which can be rewritten using exponents as (2^(1/2))^(1/3). When we multiply the exponents, we get 2^(1/6). To find out which integer this is closest to, we can look for small integers and their sixth powers to compare. The sixth power of 1 is 1 (since 1^6 = 1), and the sixth power of 2 is 64 (since 2^6 = 64). The number 2^(1/6) will be somewhere between 1 and 2, but closer to 1 since 2 is much larger when raised to the sixth power. Therefore, the integer closest to the cube root of √2 is 1.
The expression in the image is asking for the cube root of the square root of 2, written as ∛√2. This can also be written as 2^(1/3 * 1/2), which simplifies to 2^(1/6). Now, to find the integer closest to 2^(1/6), we can try to estimate this value. Since 2^(1/6) is the sixth root of 2, we're looking for a number which, when raised to the power of 6, is closest to 2. Starting with 1, we see that 1^6 = 1, which is far from 2. Next, let's take 2^6, which is 64; this is much larger than 2. Therefore, the integer closest to 2^(1/6) must be 1, since any number greater than 1 raised to the power of 6 will be greater than 2. So, the integer closest to ∛√2 is 1.
The question is asking us to find which integer is closest to the cube root of 55. To get an idea of which integers might be close, let's consider the perfect cubes near 55. The cube of 3 is \(3^3 = 27\), and the cube of 4 is \(4^3 = 64\). Since 55 lies between 27 and 64, the cube root of 55 must lie between 3 and 4. Since 55 is closer to 64 than it is to 27, the cube root of 55 will be closer to 4 than to 3. Therefore, the integer closest to the cube root of 55 is 4.
The image shows an inequality with a cube root: \[ \_ < \sqrt[3]{63} < \_ \] We need to fill in the blanks with numbers to make the inequality true. To do so, we need to find two numbers that the cube root of 63 falls between. The cube root of 63 is a bit more than the cube root of 64, which is 4, because 4^3 = 64. Since 63 is less than 64, its cube root will be slightly less than 4. Now let's find two consecutive whole numbers that the cube root of 63 will lie between. We already know that it will be less than 4 and more than 3 because 3^3 = 27 which is much less than 63. So the correct inequality is: \[ 3 < \sqrt[3]{63} < 4 \] This reads as "the cube root of 63 is more than 3 but less than 4."
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