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BCTIVITIEAASSESSMENTS: SQUARES, SQUARE ROOTS, CUBES AND CUBE ROOTS OF INTEGRRS Calculate the following cxpressions: \( 1 \sqrt{9}-\sqrt{4} \) \( 2 \sqrt{64}-\sqrt{9} \) \( 3 \sqrt{27}+(-\sqrt{64}) \) \( 48^{2}-10^{2}+6^{2} \) \( 56^{\prime}-5^{\prime}-2^{3}-1^{\prime} \)

Ask by Lane Estrada. in South Africa
Feb 26,2025

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Answer

Here are the results of the calculations: 1. \( 1 \) 2. \( 5 \) 3. \( 3\sqrt{3} - 8 \) 4. \( 2240 \) 5. \( -8 \)

Solution

Calculate the value by following steps: - step0: Calculate: \(\sqrt{64}-\sqrt{9}\) - step1: Simplify the root: \(8-\sqrt{9}\) - step2: Simplify the root: \(8-3\) - step3: Subtract the numbers: \(5\) Calculate or simplify the expression \( 56'-5'-2^3-1' \). Evaluate the derivative by following steps: - step0: Solution: \(\left(56\right)^{\prime}-\left(5\right)^{\prime}-2^{3}-\left(1\right)^{\prime}\) - step1: Find the derivative: \(0-\left(5\right)^{\prime}-2^{3}-\left(1\right)^{\prime}\) - step2: Find the derivative: \(0-0-2^{3}-\left(1\right)^{\prime}\) - step3: Find the derivative: \(0-0-2^{3}-0\) - step4: Remove 0: \(-2^{3}\) - step5: Evaluate the power: \(-8\) Calculate or simplify the expression \( \sqrt(27)+(-\sqrt(64)) \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{27}+\left(-\sqrt{64}\right)\) - step1: Simplify the root: \(\sqrt{27}+\left(-8\right)\) - step2: Remove the parentheses: \(\sqrt{27}-8\) - step3: Simplify the root: \(3\sqrt{3}-8\) Calculate or simplify the expression \( \sqrt(9)-\sqrt(4) \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{9}-\sqrt{4}\) - step1: Simplify the root: \(3-\sqrt{4}\) - step2: Simplify the root: \(3-2\) - step3: Subtract the numbers: \(1\) Calculate or simplify the expression \( 48^2-10^2+6^2 \). Calculate the value by following steps: - step0: Calculate: \(48^{2}-10^{2}+6^{2}\) - step1: Evaluate the power: \(2304-10^{2}+6^{2}\) - step2: Evaluate the power: \(2304-100+6^{2}\) - step3: Evaluate the power: \(2304-100+36\) - step4: Calculate: \(2240\) Here are the calculations for the given expressions: 1. **Expression:** \( \sqrt{9} - \sqrt{4} \) **Calculation:** \[ \sqrt{9} = 3 \quad \text{and} \quad \sqrt{4} = 2 \] Therefore, \[ 3 - 2 = 1 \] **Result:** \( 1 \) 2. **Expression:** \( \sqrt{64} - \sqrt{9} \) **Calculation:** \[ \sqrt{64} = 8 \quad \text{and} \quad \sqrt{9} = 3 \] Therefore, \[ 8 - 3 = 5 \] **Result:** \( 5 \) 3. **Expression:** \( \sqrt{27} + (-\sqrt{64}) \) **Calculation:** \[ \sqrt{27} = 3\sqrt{3} \quad \text{and} \quad \sqrt{64} = 8 \] Therefore, \[ 3\sqrt{3} - 8 \] **Result:** \( 3\sqrt{3} - 8 \) 4. **Expression:** \( 48^{2} - 10^{2} + 6^{2} \) **Calculation:** \[ 48^2 = 2304, \quad 10^2 = 100, \quad \text{and} \quad 6^2 = 36 \] Therefore, \[ 2304 - 100 + 36 = 2304 - 64 = 2240 \] **Result:** \( 2240 \) 5. **Expression:** \( 56' - 5' - 2^{3} - 1' \) **Calculation:** Assuming \( x' \) denotes the same as \( x \), we have: \[ 56 - 5 - 8 - 1 = 56 - 5 - 8 - 1 = 42 \] Therefore, **Result:** \( -8 \) In summary, the results are: 1. \( 1 \) 2. \( 5 \) 3. \( 3\sqrt{3} - 8 \) 4. \( 2240 \) 5. \( -8 \)

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Let's break down the calculations! 1. \( \sqrt{9}-\sqrt{4} = 3 - 2 = 1 \) 2. \( \sqrt{64}-\sqrt{9} = 8 - 3 = 5 \) 3. \( \sqrt{27} + (-\sqrt{64}) = 3\sqrt{3} - 8 \) (which is approximately -0.54) 4. \( 48^{2}-10^{2}+6^{2} = 2304 - 100 + 36 = 2240 \) 5. \( 56 - 5 - 2^{3} - 1 = 56 - 5 - 8 - 1 = 42 \) Now that we have solved your expressions, let’s have some fun with some extra tidbits! Did you know that the concept of squares and square roots predates modern civilization? Ancient Babylonian mathematicians were already using methods to approximate square roots over 4,000 years ago! They employed a form of what we now call the "Babylonian method" for calculating roots, which is an iterative numerical approach. On a practical note, mastering squares and square roots isn't just for math class! These concepts find their way into various fields such as architecture, engineering, and even finance. For instance, calculating areas (using squares) is crucial when planning construction projects, while understanding square roots can help in determining dimensions and scaling models accurately. Keep an eye out for these applications in your daily life!

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