300 likes | 314 Views
Explore direct, joint, inverse, and combined variation problems with real-world examples and interactive exercises. Enhance your understanding of mathematical relationships in this comprehensive lesson.
E N D
Five-Minute Check (over Lesson 8–4) CCSS Then/Now New Vocabulary Key Concept: Direct Variation Example 1: Direct Variation Key Concept: Joint Variation Example 2: Joint Variation Key Concept: Inverse Variation Example 3: Inverse Variation Example 4: Real-World Example: Write and Solve an Inverse Variation Example 5: Combined Variation Lesson Menu
A. ans B. C. D. 5-Minute Check 1
A. B. C. D. 5-Minute Check 2
A. B. C. D. 5-Minute Check 3
A. B. C. D. 5-Minute Check 4
A.x = 2 B.x = –2 C.x = –8 D.x = –4 5-Minute Check 5
Content Standards A.CED.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. Mathematical Practices 1 Make sense of problems and persevere in solving them. 4 Model with mathematics. CCSS
You wrote and graphed linear equations. • Recognize and solve direct and joint variation problems. • Recognize and solve inverse and combined variation problems. Then/Now
direct variation • constant of variation • joint variation • inverse variation • combined variation Vocabulary
Direct Variation y1 = –15, x1 = 5, and x2 = 3 Cross multiply. Direct Variation If y varies directly as x and y = –15 when x = 5, find y when x = 3. Use a proportion that relates the values. Example 1
Direct Variation –45 = 5y2 Simplify. –9 = y2 Divide each side by 5. Answer: When x = 3, the value of y is –9. Example 1
A.–28 B. C. D. If y varies directly as x and y = 12 when x = –3, find y when x = 7. Example 1
Joint Variation Suppose y varies jointly as x and z. Find y when x = 10 and z = 5, if y = 12 when x = 3 and z = 8. Use a proportion that relates the values. Joint variation y1 = 12, x1 = 3, z1 = 8,x2 = 10, and z2 = 5 Cross multiply. Example 2
Joint Variation 600 = 24y2 Simplify. 25 = y2 Divide each side by 24. Answer: When x = 10 and z = 5, y = 25. Example 2
A. B. C. D. Suppose y varies jointly as x and z. Find y when x = 3 and z = 2, if y = 11 when x = 5 and z = 22. Example 2
Inverse Variation If r varies inversely as t and r = –6 when t = 2, find r when t = –7. Inverse Variation r1 = –6, t1 = 2, and t2 = –7 Cross multiply. Example 3
Inverse Variation Simplify. Divide each side by –7. Answer: When t = –7, r is . Example 3
A. B.4 C.16 D.144 If a varies inversely as b and a = 3 when b = 8, find a when b = 6. Example 3
Write and Solve an Inverse Variation SPACEThe apparent length of an object is inversely proportional to one’s distance from the object. Earth is about 93 million miles from the Sun. Venus is about 67 million miles from the Sun. How much larger would the diameter of the Sun appear on Venus than on Earth? Let L1 = the apparent diameter of the Sun from Venus and let D1 = the distance of Venus from the Sun. Let L2 = the apparent diameter of the Sun from Earth and let D2 = the distance of Earth from the Sun. Example 4
V = Write and Solve an Inverse Variation Let the apparent diameter of the Sun from Earth equal1 unit and the apparent diameter of the Sun from Venusequal V. Use a proportion that relates the values. Original equation L1D1 = L2D2 V ● 67 = 1 ● 93 L1= V, D1 = 67, L2 = 1, D2 = 93 Divide each side by 67. V≈ 1.39 Simplify. Example 4
Write and Solve an Inverse Variation Answer: From Venus, the diameter of the Sun will appear about 1.39 times as large as it appears from Earth. Example 4
SPACE The apparent length of an object is inversely proportional to one’s distance from the object. Mars is about 142.5 million miles from the Sun and Earth is about 93 million miles away. How much smaller would the diameter of the Sun appear on Mars than on Earth? A. 13,252.50 times B. 49.50 times C. 1.53 times D. 0.65 times Example 4
Combined Variation Suppose f varies directly as g, and f varies inversely as h. Find g when f = 6 and h = –5, if g = 18 when h = 3 and f = 5. First, set up a correct proportion for the information given. g varies directly as f, so g goes in the numerator. h varies inversely as f, so h goes in the denominator. Solve for k. Example 5
Combined Variation Set the two proportions equal to each other. f1 = 5, h1 = 3, g1 = 18, f2 = 6, and h2 = –5 Cross multiply. Simplify. Divide each side by 15. Answer: When f = 6 and h = –5, the value of g is –36. Example 5
Suppose f varies directly as g, and f varies inversely as h. Find g when f = 6 and h = –16, if g = 10 when h = 4 and f = –6. A. –30 B. 30 C. 36 D. 40 Example 5