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Presentation regarding Variations (Grade 8-9)
- 1. REVIEW FOR SECOND QUARTER Prepared by: Mr Allan P. Limin
- 3. Variations
- 4. Direct Variation y = kx or k = π π Direct Square Variation y = πππ or k = π ππ Inverse Variation y = π π or k = xy Joint Variation y = kxz or k = π ππ Combined Variation y = ππ π or k = ππ π
- 6. Graph
- 9. Graph
- 10. y = kxz 20 = k (4)(3) 20 = k 12 k = π π y = kxz y = ( π π )(2)(3) y = 10
- 12. Problem: A. Write each in equation form. 1. The volume V of a cylinder varies directly as its height h. 2. The area A of a square varies directly as the square of its side s. 3. The length l of a rectangular field varies inversely as its width w. 4. The volume of cylinder V varies jointly as its height h and the square of the radius r. 5. The electrical resistance R of a wire varies directly as its length l and inversely as the square of its diameter d.
- 13. 1. V = kh 2. A = kππ 4. V = khππ 3. l = π π 5. R = ππ π π. Answers
- 14. B. Find k and express the equation of variation. 1. y varies directly as x and y=30 when x=8. 2. x varies directly as the square of y, and x=6, when y=8. 3. c varies jointly as a and b, and c=45, a=15 and b=14. 4. x varies directly as y and inversely as z, when x=15, y=20 and z=40.
- 15. 1. y = kx 30 = k8 ππ π = k y = ππ π x 2. x = kππ 6 = k(π)π 6 = k64 π ππ = k x = π ππ ππ 3. c = kab 45 = k(15)(14) 45 = k 210 π ππ = k c = π ππ ab 4. x = ππ π 15 = πππ ππ 30 = k x = πππ π
- 16. C. Solve for the indicated variable in each of the ff. 1. If y varies directly as x, and y = -18 when x = 9, find y when x = 7. 2. If y varies directly as the square of x, and y=36 when x=3, find y when x=5. 3. z varies jointly as x and y, and z=60 when x=5, y=6, find z when x=7 and y=6. 4. If r varies directly as s and inversely as the square of u, and r=2 when s=18 and u=2, find r when u=3 and s =27.
- 17. Answers!!! 1. y = kx -18 = k9 -2 = k y = (-2)(7) y = -14 2. y = kππ 36 =k(π)π 36 =k9 4 = k y = (4) (π)π y = 100 3. z = kxy 60 = k(5)(6) 60 = k 30 2 = k z = (2)(7)(6) z = 84 4. r = ππ ππ 2 = πππ (π)π π π = k r = π π (ππ) (π)π r = ππ π r = π π
- 18. Worded Problems 1. Candies are sold at 50 centavos each. How much will a bag of 420 candies cost? y = kx .50 = k1 .50 = k y = (.50)(420) y = β±210.00 ππ ππ = ππ ππ .ππ π = ππ πππ ππ = β±210.00
- 19. 2. When a body falls from rest, its distance from the starting point is directly proportional to the square of the time during which it is falling. In 2 seconds, a body falls through 19.57 meters. How far will it fall in 5 seconds? d = kππ 19.57 = k(π)π 19.57 = k4 4.8925 = k d = (4.8925) (π)π d = 122.31 ππ ππ π = ππ ππ π ππ.ππ ππ = ππ ππ πππ = 489.25 ππ = 122.31
- 20. 3. The mass of a rectangular sheet of wood varies jointly as the length and the width. When the length is 20 cm and the width is 10 cm, the mass is 200 g. Find the mass when the length is 15 cm and the width is 10 cm. m = klw 200 = k(20)(10) 200 = k200 1 = k m = (1)(15)(10) m = 150 grams ππ ππππ = ππ ππππ πππ (ππ)(ππ) = ππ (ππ)(ππ) 200ππ = 30,000 ππ = 150 grams
- 21. 4. The current I varies directly as the electromotive force E And inversely as the resistance R. If in a system a current of 20 amperes flows through a resistance of 20 ohms with an Electromotive force of 100 volts, find the current that 150 volts will send through the system. I = ππ¬ πΉ 20 = ππππ ππ 400 = k100 4 = k I = ππ¬ πΉ I = (π)πππ ππ I = 30 π°ππΉπ π¬π = π°ππΉπ π¬π (ππ)(ππ) πππ = π°πππ πππ 60,000 = π°π2,000 30 = π°π
- 23. Laws of Exponents 1. Product of Powers: ππ β ππ = ππ+π 2. Power of a Power: (ππ )π = πππ 3. Power of a Product: (ππ)π = ππ ππ 4. Power of a Monomial: (ππ ππ )π = πππ πππ 5. Quotient of Powers: ππ ππ =ππβπ
- 24. 6. Power of a Fraction: ( π π )π = ππ ππ 7. Zero Exponent: ππ = π 8. Negative Exponent: πβπ = π ππ 9. Rational Exponent: (π π π)π = π π π or π βπ π = π π π π
- 25. Evaluate: 1. ππ β ππ 2. (ππ )π 3. (ππ)π 4. (ππ ππ )π 5. ππππ πππ 6. ( π π )π 7. πππ ππ 8. πβπ 9. ππβππ ππππβπ 1. ππ 2. πππ 3. 8ππ 4. πππππ 5. 100 6. ππ ππ 7. πππ 8. π ππ 9. πππ ππ
- 26. Radicals
- 27. Definition: If a is a nonnegative real number, the nonnegative number b such that ππ = a is the principal square root of a and Is denoted by b = π.
- 29. Example:
- 30. Example:
- 31. Example:
- 32. Example:
- 35. 1. The equivalent of π 2. Simplify πππ 3. Evaluate πππππ 4. Simplify ( π ππ)βπ 5. Simplify 2 πππ + 3 πππ + 4 πππ + 2 πππ 6. Simplify ππ + ππ + ππ - 10 πππ 7. Multiply (4 π)(3 π) 8. Find the product of ( π π)( π) 9. Rationalize the denominator of ππ π π 10.Divide: πππβπππ ππππ Evaluate: π π π 3 ππ πππ 1/16 18 ππ +14 ππ -41 π 24 π π ππ 2 π ππ πππ ππ π ππ