Karnataka Board Class 10 Mathematics — Flashcards for Quick Revision

A: Step 1: Divide 140 by smallest prime (2) → 140 ÷ 2 = 70 Step 2: 70 ÷ 2 = 35 Step 3: 35 ÷ 5 = 7 Step 4: 7 is prime Therefore: 140 = 2² × 5 × 7 Answer: 140 = 2² × 5¹ × 7¹

A: Step 1: Prime factorization 96 = 2⁵ × 3¹ 404 = 2² × 101¹ Step 2: HCF = Product of lowest powers of common factors Common factor: 2 HCF = 2² = 4 Step 3: LCM = Product of highest powers of all factors L

A: Step 1: Prime factorization 26 = 2¹ × 13¹ 91 = 7¹ × 13¹ Step 2: Find HCF and LCM HCF = 13¹ = 13 (common factor) LCM = 2¹ × 7¹ × 13¹ = 182 Step 3: Verify the relationship HCF × LCM = 13 × 182 = 2366 Pr

A: Step 1: Find AC using Pythagoras theorem AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 AC = √625 = 25 cm Step 2: Apply trigonometric ratios sin A = opposite/hypotenuse = BC/AC = 7/25 = 0.28 cos A = adj

A: Step 1: Draw right triangle where opposite = 4k, adjacent = 3k Step 2: Find hypotenuse using Pythagoras hypotenuse² = (4k)² + (3k)² = 16k² + 9k² = 25k² hypotenuse = 5k Step 3: Calculate all ratios s

A: Step 1: Recall values from table sin 30° = 1/2 cos 30° = √3/2 Step 2: Square each value sin² 30° = (1/2)² = 1/4 cos² 30° = (√3/2)² = 3/4 Step 3: Add them sin² 30° + cos² 30° = 1/4 + 3/4 = 4/4 = 1 N

A: Step 1: Factor the polynomial. Look for two numbers that multiply to 6 and add to -5. Those are -2 and -3. Step 2: x² - 5x + 6 = (x - 2)(x - 3) Step 3: Set each factor to zero: x - 2 = 0 or x - 3 = 0

A: Step 1: Find zeros by factoring: 2x² - 7x + 3 = (2x - 1)(x - 3) Zeros: x = 1/2 and x = 3 Step 2: Check sum of zeros: Sum = 1/2 + 3 = 7/2 From formula: -b/a = -(-7)/2 = 7/2 ✓ Step 3: Check product of

A: Given: p(x) = x² - 6x + 8 Compare with ax² + bx + c where a = 1, b = -6, c = 8 Step 1: Sum of zeros formula: α + β = -b/a = -(-6)/1 = 6 Step 2: Product of zeros formula: αβ = c/a = 8/1 = 8 Answer: