Just Solving Questions

🧮 Most Useful Modular Arithmetic Property to Simplify Calculations Question: What is a simple modular arithmetic property you use most often to quickly simplify a calculation? Modular arithmetic makes working with large numbers easier by focusing only on their remainders. Among its many properties, one stands out as the most practical and frequently used for quick  simplifications. 🔹 The Core Property The most useful property is: (a + b) mod m = ((a mod m) + (b mod m)) mod m (a × b) mod m = ((a mod m) × (b mod m)) mod m This means you can take the remainder (mod) of numbers as you go — without having to calculate the full product or sum first. 💡 Why It Works Instead of handling large numbers directly, modular arithmetic lets you reduce them early. This keeps calculations smaller, faster, and error-free. ⚙️ Example Find (2347 × 6789) mod 10 (2347 mod 10) = 7 (6789 mod 10) = 9 (7 × 9) mod 10 = 63 mod 10 = 3 ✅ Final Answer: 3 You simplified the whole prob...

🧮 Simplify 0.027⁻¹/³ Easily | Step-by-Step Cube Root & Exponent Guide Question: Simplify 0.027⁻¹/³ Let’s break it down step-by-step with a simple explanation. This concept often appears in algebra and exponent exercises. 🔹 Step 1: Understand the Negative Exponent A negative exponent means take the reciprocal of the base raised to the positive power. 0.027⁻¹/³ = 1 / 0.027¹/³ 🔹 Step 2: Find the Cube Root We can rewrite 0.027 as a fraction: 0.027 = 27 / 1000 Now, take the cube root of both numerator and denominator: (27/1000)¹/³ = 3/10 = 0.3 🔹 Step 3: Apply the Reciprocal Since we had a negative exponent, we now take the reciprocal of 0.3: 1 / 0.3 = 10 / 3 Step-by-step breakdown of 0.027⁻¹/³ simplification showing reciprocal and cube root method. ✅ Final Answer: 0.027⁻¹/³ = 10/3 or 3.333… 📘 Quick Recap of Concepts a⁻ⁿ = 1 / aⁿ a¹/³ means the cube root of a. Find the cube root first, then take the reciprocal for negative powers. 💡...

How to Internalize Math and Physics Concepts: A Step-by-Step Guide In the journey of learning, there’s a difference between knowing a formula and truly internalizing a concept. Especially in fields like mathematics and physics, internalization means the idea becomes part of how you think, rather than something you just recall for an exam. Below is a research-informed, step-by-step guide to help you internalize math and physics concepts, build mastery and retention, and apply them fluently. 1. Start with active recall, not passive reading Research shows that attempting problems before reviewing your notes boosts long-term retention. Students who try to solve problems first and then refer to notes actually learn more deeply.  So, in practice: after a lecture or reading session, pick a problem quickly (10-15 minutes) and try to wrestle with it from memory. Write down your thinking, even if you guess. Then check your resources and fill gaps. This practice builds neural connectio...

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 Find all the roots of (1+z)^5=(1-z)^5 | Complex Analysis 1. Simplify the equation: Divide both sides by (1 - z)^5: (1 + z)^5 / (1 - z)^5 = 1 Recognize this as a special case: This equation is of the form (a/b)^5 = 1, where a = 1 + z and b = 1 - z. 2. Find the fifth roots of unity: The equation (a/b)^5 = 1 implies that a/b is a fifth root of unity. The fifth roots of unity are given by: e^(2kπi/5), where k = 0, 1, 2, 3, 4 3. Express a/b in terms of fifth roots of unity: a/b = e^(2kπi/5) 4. Solve for z: Substitute a = 1 + z and b = 1 - z: (1 + z) / (1 - z) = e^(2kπi/5) Cross-multiply: 1 + z = e^(2kπi/5) * (1 - z) Expand and rearrange: z + z*e^(2kπi/5) = e^(2kπi/5) - 1 Factor out z: z * (1 + e^(2kπi/5)) = e^(2kπi/5) - 1 Solve for z: z = (e^(2kπi/5) - 1) / (1 + e^(2kπi/5)) 5. Simplify the expression for z: Multiply the numerator and denominator by e^(-kπi/5): z = (e^(kπi/5) - e^(-kπi/5)) / (e^(kπi/5) + e^(-kπi/5)) Use Euler's formula: z = (2i sin(kπ/...

Q) Suppose L:R3to R3 is a linear operator and L[1,0,0]=[-2,1,0], L=[0,1,0]=[-3,-2,1], L[0,0,1]=[0,-1,3]  find basis Find L[-3,2,4]. give a formula for L[x,y,z] for all [x,y,z] belongs to R3 Given that L L L is a linear operator and L [ 1 , 0 , 0 ] = [ − 2 , 1 , 0 ] L[1, 0, 0] = [-2, 1, 0] L[1,0,0]=[−2,1,0], L [ 0 , 1 , 0 ] = [ − 3 , − 2 , 1 ] L[0, 1, 0] = [-3, -2, 1] L[0,1,0]=[−3,−2,1], and L [ 0 , 0 , 1 ] = [ 0 , − 1 , 3 ] L[0, 0, 1] = [0, -1, 3] L[0,0,1]=[0,−1,3], we can write: [ − 3 , 2 , 4 ] = − 3 [ 1 , 0 , 0 ] + 2 [ 0 , 1 , 0 ] + 4 [ 0 , 0 , 1 ] = − 3 L [ 1 , 0 , 0 ] + 2 L [ 0 , 1 , 0 ] + 4 L [ 0 , 0 , 1 ] \begin{align*} [-3, 2, 4] &= -3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1] \\ &= -3L[1, 0, 0] + 2L[0, 1, 0] + 4L[0, 0, 1] \end{align*} [−3,2,4]​=−3[1,0,0]+2[0,1,0]+4[0,0,1]=−3L[1,0,0]+2L[0,1,0]+4L[0,0,1]​ Using the given values for L [ 1 , 0 , 0 ] L[1, 0, 0] L[1,0,0], L [ 0 , 1 , 0 ] L[0, 1, 0] L[0,1,0], and L [ 0 , 0 , 1 ] L[0, 0, 1] L[0,0,1], we can compute: − 3 L [ 1 , 0 ...

Q) Suppose L:R3-> R3 is a linear operator  and L([1,0,0])=[-3,2,4], l=([0,1,0]) = [5,-1,3] and L([0,0,1])=[-4,0,-2]find L([6,2,-7]) and find L([x,y,z]) for any [x,y,z] belongs to R3. To find L ( [ 6 , 2 , − 7 ] ) L([6,2,-7]) L([6,2,−7]), we'll use linearity: Given: L ( [ 1 , 0 , 0 ] ) = [ − 3 , 2 , 4 ] L([1,0,0]) = [-3,2,4] L([1,0,0])=[−3,2,4] L ( [ 0 , 1 , 0 ] ) = [ 5 , − 1 , 3 ] L([0,1,0]) = [5,-1,3] L([0,1,0])=[5,−1,3] L ( [ 0 , 0 , 1 ] ) = [ − 4 , 0 , − 2 ] L([0,0,1]) = [-4,0,-2] L([0,0,1])=[−4,0,−2] We have [ 6 , 2 , − 7 ] = 6 [ 1 , 0 , 0 ] + 2 [ 0 , 1 , 0 ] − 7 [ 0 , 0 , 1 ] [6,2,-7] = 6[1,0,0] + 2[0,1,0] - 7[0,0,1] [6,2,−7]=6[1,0,0]+2[0,1,0]−7[0,0,1]. Since L L L is linear, we can apply it to each term separately and then add the results: L ( [ 6 , 2 , − 7 ] ) = 6 L ( [ 1 , 0 , 0 ] ) + 2 L ( [ 0 , 1 , 0 ] ) − 7 L ( [ 0 , 0 , 1 ] ) L([6,2,-7]) = 6L([1,0,0]) + 2L([0,1,0]) - 7L([0,0,1]) L([6,2,−7])=6L([1,0,0])+2L([0,1,0])−7L([0,0,1]) = 6 [ − 3 , 2 , 4 ] + 2 [ 5 , − 1 , 3 ] −...