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Formulas

Evaluate formulas with given values and solve algebraic problems. Includes distance, perimeter, area calculations, and solving equations. Practice exercises and solutions provided.

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Formulas

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  1. Formulas ALGEBRA 1 LESSON 2-6 (For help, go to Lessons 1-2, 1-4 and 1-6.) Evaluate each formula for the values given. 1. distance: d = rt, when r = 60 mi/h and t = 3 h 2. perimeter of a rectangle: P = 2 + 2w, when = 11 cm, and w = 5 cm 3. area of a triangle: A = bh, when b = 8 m and h = 7 m 1 2 2-6

  2. Formulas ALGEBRA 1 LESSON 2-6 Solutions 1.d = rt = 60(3) = 180 mi 2.P = 2 + 2w = 2(11) + 2(5) = 22 + 10 = 32 cm 3.A = bh = (8)(7) = 4(7) = 28 m2 1 2 1 2 2-6

  3. V = wh V wh / = Divide each side by , = 0. V = wh Simplify. V h wh h = Divide each side by h, h = 0, to get w alone on one side of the equation. / V h = w Simplify. Formulas ALGEBRA 1 LESSON 2-6 Solve the formula for the volume of a rectangular prism V = wh for width w in terms of its volume V, length , and its height h. 2-6

  4. y + 34 4x4 = Divide each side by 4. y + 34 = xSimplify. Formulas ALGEBRA 1 LESSON 2-6 Solve y = 4x – 3 for x. y + 3 = 4x – 3 + 3 Add 3 to each side. y + 3 = 4xSimplify. 2-6

  5. –br–r p – z–r = Divide each side by –r. p – z–r b = –Simplify. Formulas ALGEBRA 1 LESSON 2-6 Solve z – br = p for b in terms of z, r, and p. z – br – z = p – zSubtract z from each side. –br = p – zCombine like terms. 2-6

  6. Solve for C. K = C + 273.15 Find C when K = 400. 400– 273.15 = CSubstitute 400 for K. 126.85 = C Formulas ALGEBRA 1 LESSON 2-6 The formula K = C + 273.15 gives the Kelvin temperature K in terms of the Celsius temperature C. Transform the formula to find Celsius temperature in terms of Kelvin temperature. Then, find the Celsius temperature when the Kelvin temperature is 400°. K – 273.15 = C + 273.15 – 273.15 Subtract 273.15 from each side. K– 273.15 = CSimplify. 400º K is 126.85º C. 2-6

  7. Solve each equation for the given variable. 1. 3x + 2y = z; y2. = ; a 3. 3x + 4 = 2(3 – y); y 4. The formula v2 = can be used to find the constant speed v of a satellite revolving around Earth in a circular orbit of radius r. G is a number known as the gravitational constant and M is the mass of Earth. Transform this formula to find the mass of Earth. 1 2 3 2 cd 3 a – b c d 3 y = z – x a = + b 3 2 y = – x + 1 GM r v2r G M = Formulas ALGEBRA 1 LESSON 2-6 2-6

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