Karnataka Board Class 9 Mathematics — Flashcards for Quick Revision

A: Step 1: Divide 7 by 8 using long division: 0.875 8)7.000 64 --- 60 56 --- 40 40 --- 0 Step 2: Since remainder becomes 0, it's a terminating decimal. Answer: 7/8 = 0.875 (terminating

A: Step 1: Let x = 0.333... Step 2: Multiply by 10: 10x = 3.333... Step 3: Subtract original equation: 10x - x = 3.333... - 0.333... Step 4: Simplify: 9x = 3 Step 5: Solve for x: x = 3/9 = 1/3 Answer: 0

A: Step 1: Find the value of √25 √25 = 5 (since 5² = 25) Step 2: Check if 5 can be written as p/q 5 = 5/1, where p = 5, q = 1 (both integers, q ≠ 0) Step 3: Conclusion Since √25 = 5 can be expressed as

A: Step 1: Identify triangle type → AB = AC, so triangle ABC is isosceles. Step 2: Apply isosceles triangle property → Angles opposite to equal sides are equal, so ∠B = ∠C. Step 3: Let ∠B = ∠C = x. Step

A: Step 1: Identify given information → AB = DE = 4 cm, BC = EF = 3 cm, ∠B = ∠E = 90°. Step 2: Check congruence rule → We have two sides and included angle. Step 3: Apply SAS rule → In △ABC and △DEF: AB

A: Step 1: Identify triangle type → PQ = QR = 7 cm, so triangle PQR is isosceles. Step 2: Property of isosceles triangle → Altitude from vertex angle bisects the base. Step 3: Since QS is altitude to PR,

A: Step 1: Identify the highest power of x → x³ has power 3. Step 2: Degree = 3. Step 3: Since degree = 3, it's a cubic polynomial. Step 4: Count terms → 4 terms, so it's not a monomial, binomial, or tri

A: For p(2): Step 1: Substitute x = 2 → p(2) = 3(2)² - 5(2) + 2. Step 2: Calculate → p(2) = 3(4) - 10 + 2 = 12 - 10 + 2 = 4. For p(-1): Step 1: Substitute x = -1 → p(-1) = 3(-1)² - 5(-1) + 2. Step 2: Cal

A: Step 1: Substitute x = 3 in p(x). Step 2: p(3) = (3)² - 5(3) + 6. Step 3: Calculate → p(3) = 9 - 15 + 6 = 0. Step 4: Since p(3) = 0, x = 3 is a zero of the polynomial. Answer: Yes, x = 3 is a zero bec