MATH 2 part 1

k = (27)(3)k = 81The constant of variation is 8, the equation of variation is y = 81 x To find y when x = 9 :y = 81 xy = 81 9y=95. An inverse variation is given by xy = k , where k is constant. If x = 15 when y =9: a) what is the value of k ? b) find the value of y when x = 6 .Solution: by substitution a) xy = k (15)(9) = k k = 135 b) since xy = k , then y = k xThe equation of the variation is y = 135 x To find y when x = 6 : y = 135 x y = 135 6 y = 22 1 2 5

6. If y varies inversely as x and y = 4 when x = 9, find y when x =12. Consider again the equation of inverse variation: xy = k , k ≠ 0If ( x1, y1 ) and ( x2 , y2 ) are any two solutions, then x1 y1 = k equation (1) x2 y2 = k equation (2)Solving equations (1) and (2) for k gives x1 y1 = x2 y2 or x1 = y2 x2 y1Solution: Let x1 = 9 , y1 = 4 , x2 = 12 , find y2 x1 y1 = x2 y2 (9)(4) = (12)( y2 ) by substitution y2 = 36 12 y2 = 3 Then y = 3 when x = 12.7. If y varies inversely as x and y = 10 when x = 2, find y when x = 10.You may choose from two solutions:Solution 1: First Set the relation, and then find the constant of variation k . xy = k (2)(10) = k k = 20The equation of variation is y = 20 x 6

Find y when x = 10 by substitution: y = 20 x y = 20 10 y=2 Solution 2: Use the equation x1 y1 = x2 y2 Let x1 = 2 , y1 = 10 , x2 = 10 , find y2 By substitution: 2(10) = 10(y2) 20 = 10y2 y2 = 20 10 y2 = 2 Hence y = 2 when x = 10.Try this outA. Determine if the equation expresses an inverse variation between the variables. If so, find the constant of variation. Recall that k = yx. 1. a = 10b 2. 5st = 12 3. mn = 12 4. 6m = 4 n 5. y = 10 x 7

B. Find an equation of variation where y varies inversely as x. 6. y = 25 when x = 3 7. y = 7 when x = 10 8. y = 42 when x = 25 9. y = 0.4 when x = 0.6 10. y = 45 when x = 2C. In each of the following, find the constant of variation and then solve for the indicated variable. 1. y varies inversely as x. If y = 3 when x = 4, find y when x = 6. 2. If w varies inversely as y. If w = 2 when y = 3, find w when y = 6. 3. If y varies inversely as x, and y is 10 when x = 5, find y when x = 7. 4. If w varies inversely as x and is 10 when x = 5 find w when x = 3. 5. If y varies inversely as x and y = 10 when x = 5, find y when x = 15. Lesson 2 Graph that shows inverse variation You have seen how one variable can vary directly as another variable indifferent ways. You have also seen that the graph of direct variation with the useof suitable axes is always a straight line. There are however, situations when twovariables do not vary directly but inversely. This module will lead you to another view of the graph of a variation, whichis the graph of an inverse variation. For instance, the formula for distance is d = rt. If the distance remainsfixed or constant, how will the rate and time be related?Illustration: Suppose a teacher lives 40 km away from school where she teaches. How long she takes to reach school depends on her average speed. Some possible speeds and the length of time she needs for each speed are: 8

speed in kph∗ 40 50 60 70 80 time in hours 14 2 41 5372 How do the speed and time of travel affect each other? The speed variesinversely as the time spent in traveling so that the constant of the variation isk = 40.To see clearly the relation of the two quantities, the graph of the relation is 90 1/5 2/5 3/5 4/5 1 1 1/5 80 time (hours) 70 60 Velocity 50 (kph) 40 30 20 10 0 0 You will notice that the graph is not a straight line since d = rt is not alinear equation, when any of the variables d , r , or t is given a constant value.The graph is asymptotic with the lines y = 0 and x = 0 , that is the graph of aninverse variation is a curve line that does not cross the x and y axis.Examples:1. Complete the table of values on rectangles whose area is 36 sq. m when the length and width are given then graph using a suitable interval on the axes. When l is 18 12 9 6 _ The value of w is 2 3 _Solution: The product of the length l and the width w in any of the pairs is equal to 36. Therefore, 36 is the variation constant. For a length of 9, the width must be 4 and 6 for the length of 6.∗ kilometers per hour 9

40 35 30 25 length (metres) 20 15 10 5 0 width (metres)2. The following table describes the relation of the length and the width of a rectangle with constant area. Graph the table.Width (x) 2 3 4 6 8Length (y) 6 4 3 2 3 2length 7 6 5 4 3 2 1 0 0123456789 width 10

3. The given tables are inverse variation. Write the equation of the given relations and determine the constant of variation. a. w 1 2 3 4 6 l 48 24 16 12 8 b. a 2 4 5 b 3 3/2 6/5 Solution: a. The equation of the relation is l = k , k = 48 w b. The equation of the relation is b = k , k = 6 a4. Given the graph of inverse variation; y is inversely proportional to x , find constant of variation and the equation. 4.50 4.00 3.50 3.00 2.50 2.00 1.50 1.00 0.50 0.00 0123456 11

Solution: The constant of variation: xy = k (1)(4) = k 4=kThe equation: y=4 xTry this outA. Graph the following tables, state whether the graph illustrates an inverse variation. If it does, determine the constant of variation.1. x 1 2 3 6 y 3 31 1 222. x 1 2 3 4 y 2 8 18 323. x 2 3 4 6 y64324. x 3 6 9 18y632 15. x 2 3 4 5 y 4 6 8 10 12

Lesson 3 Problems involving inverse variation Now, that you have gained analytical concept on how to deal with inversevariation, this lesson will enhance your problem solving skill in dealing withsituations about inverse variation.Examples: 1. The number of hours required to finish a certain job varies inversely as the number of persons on the job. If 9 persons require 10 hours to finish the job, how long should it take for 30 persons to finish the job? Solution: Let t - number of hours required to finish the job n - number of people on this job. The equation: t = k n Solving for k , from the first set of given data, 10 = k 9 k = (9)(10) k = 90 To find the time t , 30 persons should take to finish the job, t = 90 n t = 90 30 t = 3 hours Alternative solution: You can use t1n1 = t2n2 (see lesson 1, example 5), where t2 is the required time for 30 persons to finish the job. 13

t1n1 = t2n2 (10)(9) = t2 (30) t2 = 90 30 t2 = 32. The bases of triangles are inversely proportional to their altitudes. The base of a certain triangle is 14 cm and its altitude is 20 cm. Find the base of another triangle having the same area as the first, whose altitude is 28 cm.Solution: b - base of the triangle let a - altitude k - areathe equation: substitute the given data b= k a 14 = k 20 k = (20)(14) k = 280To find the base b where the altitude is 28 cm: b = 280 a b = 280 28 b = 10Alternative solution: By proportion, you have a1b1 = a2b2 where b2 is unknown (20)(14) = (28)b2 by substitution 14

b2 = (20)(14) 28 b2 = 10 cm3. The intensity I of light from the bulb is inversely proportional to the square of the distance s from the bulb. If I = 45 w/m2 when the distance s = 10 m. Find the intensity of the light from the bulb at a distance of 5 m.Solution:The equation: I = k s2 45 = k substitute the given data (10)2 k = (45)(10)2 k = (45)(100) k = 4500Finding I when s = 5 m: I = 4500 s2 I = 4500 52 I = 4500 25 I = 180 w/m2Alternative solution: By proportion, you have I1s12 = I2s22 where I2 is unknown (45)(10)2 = I2 (5)2 by substitution I2 = (45)(100) 25 I2 = 180 w/m2 15

4. The speed of a gear wheel varies inversely as the number of teeth of the gear. If a gear wheel which has 48 teeth makes 3 revolutions per minute. How many revolutions per minute will a gear wheel with 96 teeth make?Solution: s = speed of a gear ( rev/min) let n = number of teeth of the gearThe equation: s=k substitute the given data n 32 = k 48 k = (48)(32) k = 1536 Solving for the number of revolutions per minute of a gear with 96teeth can make, s = 1536 n s = 1536 96 s = 16 revolutions per minute (rpm)Try this outA. Solve the following. You may use either of the solutions presented to you.1. A pyramid with a square base has a fixed volume. The height h varies inversely as the area of the base s2 :a. What is the constant of proportionality if h = 8 , when side s = 6 ?b. If h = 8 , when side s = 6 , what is h when s = 4 ?2. The frequency f or number of vibrations of a string under constant tension is inversely proportional to the length l of the string. If a 48 cm string vibrates 192 cycles per second:a. Find the constant of proportionalityb. Find the number of vibrations per second of a 32 cm string will make. 16

B. How do you pronounce VOLIX ?Answer the questions then decode the answerThe number of kg W of water in a The amount C that a family spends onhuman body varies directly as the total car expenses varies directly as itsweight B. A person weighing 75 kg income I. A family makingcontains 54 kg of water. How many kg P 21,760 a year will spend P3264 aof water are in a person weighing 95 year for car expenses. How much willkg? a family making P 30,000 a year spend for car expenses. m vA production line produces 15 compactdisc player every 8 hrs. How many players It takes 16 hrs for 2 people to resurfacecan it produce in 37 hrs? a gym floor. How long will it take 6 people to do the job? e iThe time t required to drive a fixed The volume V of a gas varies inverselydistance varies inversely as the speed r. as the pressure P on it. The volume ofIt takes 5 hrs at 55 mph to drive a fixed a gas is 200 cubic cm (cm3) under adistance. How long would it take at 40 pressure of 32 kg/cm2. What will be itsmph? volume under a pressure of 20 kg/cm2? n nIt takes 4 hrs for 9 cooks to prepare a The height H of triangle of fixed areaschool lunch. How long will it take 8 varies inversely as the base B. Supposecooks to prepare the lunch? the height is 59 cm when the base is 40 cm. Find the height when the base is 8 o cm. eA person works for 15 hrs and makes The current I in an electrical conductorP7125. How much will the person varies inversely as the resistance R ofmake by working 35 hrs? the conductor. The current is 2 amperes when the resistance is 960 u ohms. What is the current when the resistance is 540 ohms? lI= P2000 000 3 5/9 hr I= 3.56 lbs P 2625 68.4 kg. 25 cmt= 6.875 48 hrs 320 cm3 19.73 17

Let ‘s summarizeDefinition:If two variables x and y are so related that their product is a non-zero constant, that y = k or k = xy , then y is said to vary inversely as x , or xy is inversely proportional to x as x1 = y2 or x1 y1 = x2 y2 x2 y1As for direct variation, y may vary inversely with the nth power of x ,provided that n > 0 . The general equation for an inverse variation or inverseproportion is y = k xn What have you learnedI. Direction: Fill in the blanks with either directly or inversely to make the statement true. 1. A man deposited in his savings account a fixed sum s every month. At the end of n months he has saved P 3000, then n varies___________ as s . 2. A rectangle is 30 cm long. If its width is w and its area is A, then A varies _____________________ as w. 3. A polygon has n sides and each side is 6cm long. The perimeter p varies ____________________ as n. 4. The number of days that will be required to dig a ditch varies____________________ as the number of men who will dig the ditch. 5. The cost of mangoes varies ________________ as the price of the mango. 18

6. The number of chicos I can get for P 100 varies ______________ as the price per kilo of chico.7. A group of businessmen sponsor a project that will cost P 100,000. the share of each member varies __________________ as the number of businessmen in the group.8. A sidewalk 500 meters long is made up a certain number of (n) of stone slabs each having a length (l). then n varies ____________ as l.9. Given a certain amount of money (m), the number of ballpens (n) that can be bought varies ___________________ as the cost ( c ) of each ballpen.10. The perimeter P of a square varies ______________as the length of its sides.II. Solve for the following. Solve first for k and the equation of variation.1. If t varies inversely as l , then we can write the equation asa. t = k b. tl = k c. t = kl d. tk = d l2. Which equation shows that m varies inversely as n ?a. m = kn b. mn = k c. n = k d. mk = n m3. y varies inversely as x. If y = 9 when x = 1/3, find x when y = 8.a. 3 b. 8 c. 3 d. 8 8 34. y varies inversely as the square root of x. If the point (4,2) is on the graphof the variation, find x when y = 4.a. 2 b. 4 c. 6 d. 85. y varies inversely as x. If y = 60 when x = 12, what is the constant ofvariation?a. 72 b. 720 c. 20 d. 76. y varies inversely as the square of r . If y = 8 when r = 1 . Find y when 4r = 3. 2a. 1/9 b. 2/9 c. 1/3 d. 4/9 e. none of these 19

Answer KeyA. How much do you know. 1. increases, increases 2. decreases, increases 3. increases, decreases 4., increases, decreases 5. increases, increasesB. 1. inverse 2. inverse 3. inverse 4. no 5. noTry this outLesson 1A. 1. yes, k = 10 2. no 3. yes, k = 12 4. no 5. yes, k =10B. 6. y = 75 x 7. y = 70 x 8. y = 1050 x 9. y = 0.24 x 10. y = 90 xC. 1. k = 12, y = 2 2. k = 6, w = 1 3. k = 50, y = 7.14 4. k = 50, w = 16.67 5. k = 50, y = 3 1/3 20

y-axislesson 2A. Graph the following tables . . . 5 61. x 1 2 3 6 y 3 31 1 22inverse, k = 3 3.5 3 2.5 2 1.5 1 0.5 0 1234 x-axis2. x 1 234 y 2 8 18 32not inverse 23 x-axisy-axis 35 30 25 4 20 15 10 5 0 1 21

3. x 2 3 4 6 y6432 inverse, k =12 7 6 5 4 y 3 2 1 0 01234567 x4. x 3 6 9 18 y6321 inverse, k = 18 7 6 5 4 y 3 2 1 0 0 2 4 6 8 10 12 14 16 18 20 x 22

5. x 2 3 4 5 y468 10 direct variation, k = 1 2 12 10 8 y6 4 2 0 0123456 xLesson 3A.1. a. h = k s2 8 = k (6)2 k = (36)(8) k = 288 b. h = k s2 h = 288 (4)2 h = 182. a. k = 9216 b. f = 288 cycles per second 23

B. o lu m e 3 5/9 hr 25 cm v I= 3.56 lbs P 2625 68.4 kg. I= P2000 000 n in e t= 6.875 48 hrs 320 cm3 19.73What have you learnedI. 1. directly 2. directly 3. directly 4. inversely 5. directly 6. inversely 7. inversely 8. inversely 9. inversely 10. directlyII. 1. b 2. b or c 3. a 4. e 5. b 6. b 24

Module Integral Exponents What this module is about This module is about algebraic expression with positive, zero and negativeexponents. Here you will develop skills in rewriting algebraic expressions with zero andnegative exponents and learn to apply this in solving problems. What you are expected to learn This module is designed for you to demonstrate understanding of expressionswith positive, negative and zero exponents, and: 1. evaluate expressions involving integral exponents, 2. rewrite algebraic expressions with zero and negative exponents, 3. solve exponential equations, and 4. solve problems involving expressions with exponentsHow much do you knowA. Simplify: 6. (x2)4 7. 5a2b · 2a5b2 1. 32 · 32 8. (3ab)(2bc)(abc) 2. (52)2 9. 20x10 y7 3. 42 · 43 · 22 · 23 4. 28 5x6 y2 25 10. 20x10 y7 5. 36 5x6 y2 34

B. Evaluate the following expressions1. x0 6. –12b02. 1000 7. (-3)2(3)0 8.  a 03. 5x0 b4. (25)0 9. (3xy)05. 6 + b0 10. 3x0 a0C. Write the following expressions with the negative exponents1. 3-1 6. x −42. 10-1 y −43. 8-34. 1-4 7. (xy)-45. x-3y3 8. (4-1)-3 9.  x  −3 y ( )10. a −4 3 b 4 b −2D. Solve for x in the following equations.1. 4x = 642. 10x = 10003. 3x = 274. 23x = 645. 3x-1 = 243Solve the following problems involving exponents.6. The speed of sound is about 5.13 x 103 per second. Find the distance traveled by sound in one hour.7. After 64 days an amoeba will have approximately reproduced 1.845 x 1019 amoebas. Express the number of amoebas in standard form.8. The speed of light is 1.86 x 105 miles per second. Find the speed of light in km/sec. (1km = .62 mi) 2

9. One mole of hydrogen molecules has a mass of 2.016 g and contains 6.02 x 1023 hydrogen molecules. What is the hydrogen molecules content of a 102 mole?10. Physicist measuring the quantity of electric charge in coulombs(c) found that one coulomb is equal to 6.24 x 1018 electrons. How many electrons are there in 103 coulombs?What you will do Lesson 1 Evaluation of Expressions with Integral Exponents For all real numbers a, and integers m and n, the following laws of exponentapplies to expressions having the same base:a. Multiplication: am · an = am + nb. Powers of Power (am)n = amn.Examples:1. 32 · 33 = 32 + 3 Add the exponents = 35 35 in factored form = 3·3·3·3·3 = 243 Get the product of the exponents 26 in factored form2. (23)2 = 23(2) = 26 Add exponents = 2·2·2·2·2·2 Add exponents of the same base = 163. a4 · a-3 = a4 + (-3) = a1 or a4. 3p2q · 2p3q2 = 6p2+3q1+2 = 6 p5q3 3

5. (a5b3)2 = a10b6 Get the product of the exponentsc. Division: am ÷ an = am - nExamples:1. 75 ÷ 73 = 75 – 3 Subtract the exponent = 72 Express as factors = 7·7 = 49 The quotient2. 12a6 ÷ 4a-3 = 3a6 – (-3) Divide the numerical coefficients and= 3a9 subtract the exponents.3. x5y4 ÷ x3y =x5 – 3 y4 – 1 Subtract the exponents of = x2y3 expressions having the same base.Try this out 6. (22)3 7. (32)2 · 43A. Evaluate the following: 8. (22)2 · (3)2 9. (4 · 2)3 1. 42 · 43 10. 102 · 53 2. 5 · 52 3. 24 · 23 · 32 · 34 6. (b6)5 4. 102 · 103 7. (22b3)2 5. (6)(62)(62) 8. (5x5y3)3 9. (4c5d3e)2B. Simplify: 10. (3a3b2c )2 1. (82)2 6. r7 ÷ r6 2. (7-1)-2 7. 14a10 ÷ 2a-5 3. (42)2 8. 18x6y3 ÷ 3x3y 4. (23 · 23)2 9. 15x3y4 ÷ 5x3y 5. (54 · 5-3)2 10. 24a7b5 c3 ÷ 4a3b3c3C. Find the quotient: 1. 105 ÷ 103 2. 93 ÷ 9 3. 52 ÷ 5-2 4. 7-1 ÷ 7-2 5. 45 ÷ 4 4

Lesson 2 Zero and Negative ExponentsZero Exponents: In the quotient rule, to divide expressions with the same base, keep the base andsubtract the exponents.That is a m = a m−n anNow, suppose that you allow m to equal n. Then you have,am = am−m = a0 Equation 1amBut you know that it is also true thatam =1 Equation 2am If you compare equations (1) and (2), you can see that the following definition isreasonable.The Zero Exponent For any real number a where a ≠ 0, a0 = 1Examples:Use the above definition to simplify each expression.1. 170 = 1 any number whose exponent is 0 is equal to 1. In 6x0, the exponent 0 applies only to x.2. 6x0 = 6(1) = 6 Distributive property3. (a3b2)0 = (1)(1) = 1 In -3y0, the exponent 0 applies only to y.4. –3y0 = -3(y0) = -3(1) = -3 5

Negative Exponents: In the product rule, to multiply expressions with the same base, keep the baseand add the exponents.That is, am • an = am+n Now, suppose that you allow one of the exponents to be negative and apply theproduct rule, then you have, suppose that m = 3 and n = -3.Then, am • an = a3 • a-3 = a3 + (-3) = a0 = 1So, a3 • a-3 = 1 Divide both sides by a3,We get a −3 = 1 a3Therefore we have this definitionNegative Exponents: For any nonzero real number a and whole number n, a−n = 1 an and a-n is the multiplicative inverse/ reciprocal of an.Examples:Simplify the following expressions.(Note: To simplify will mean to write the expression with positive exponents)1. y-5 = 1 you get the reciprocal y5 you get the reciprocal and simplify you get the reciprocal and simplify2. 4-2 = 1 = 1 = 1 45 (4)(4) 163. (-3)-3 = 1 = 1 = 1 (−3)3 (−3)(−3)(−3) 27 6

4.  2 −3 = 1 = 1 = 1 Get the reciprocal to have a 3  2 3  2  2  2  8 positive exponent, simplify then 3  3  3  3  27 divide. 1 ÷ 8 = 1• 27 = 27 To divide fractions, get the 27 8 8 reciprocal and proceed to multiplication.5. 2x-3 = 2 • 1 = 2 x3 x3 The exponent –3 applies only to the variable x, and not to the coefficient 2.Caution: The expression 4w-2 and (4w)-2 are not the same. Do you see why?6. 4w-2 = 4 • 1 = 4 w2 w27. (4w)-2 = 1 = 1 = 1 (4w)2 (4w)(4w) 16w2 Suppose that a variable with a negative exponent appears in the denominator ofan expression.8. 1 = 1 Change the negative exponent in a −2 1 the denominator to positive a2 To divide, get the reciprocal of 1 a2 1• a2 = a2 1 and multiply.Try this out 6. x0y 7. (xyz)0A. Simplify each expression. 8. 10000 1. 250 9. (ab)0 2. (m4n4)0 10. –5a0 3. 8m0 4. –7t0 5. pq0 7

B. Simplify each of the following expressions.1. a-10 6. 10x-52. 2-4 7. (2y)-43. (-4)-2 8. –5t-24.  5 −2 9.  1  2  x −4 5. 3w-4 10.  2 2   3a −  Lesson 3 Solving Exponential Equations Exponential Equations are equations such that the unknown is an exponent.Solutions may be rational or irrational. The method of getting the solution is to equatethe exponents of numbers which may have the same base.Examples:Solve for x in the following equations:1. 3x = 81 Express the right-hand member as a power of 3, 81 3x = 34 =3•3•3•3 So, you have both sides as a power of the base.x=4 Cancel the base, equate the exponents, therefore x = 4.2. 5x = 125 Express 125 as a power of 5, 125 = 5 • 5 • 5 5x = 53 Equate the exponents and cancel the base x=3 Therefore, x = 3.3. 42x + 2 = 44 Both sides have the same base. 2x + 4 = 4 Cancel the base, equate the exponents. 8

2x = 2 Simplify x=1 Therefore, x = 1. Express both sides as a power of 3, 9 = 3 • 3 and4. 9x + 1 = 27x 27 = 3 • 3 •3. 32(x+1) = 33(x) Distributive Property of Multiplication. 32x + 2 = 33x Simplify 2x + 2 = 3x -x = -2 Therefore, x = 2. x=2Try this outA. Solve for x in the following equations.1. 5x = 625 6. 33x – 1 = 2432. 8x = 2 7. 2x + 2 = 323. 4x = 1 8. 23x – 3 = 64 64 9. 52x – 3 = 1254. 10x = 100005. 23x = 64 10. 2x + 3 – 23 = 1 4B. Solve for x.1. 7x = 49 6. 10x + 52 = 1252. 4-2x = 16 7. 72x – 1 = 7213. 83x – 1 = 820 8. 3-3x = 14. 62x – 4 = 616 27 9. 16x – 1 = 4x – 45. 4x + 162 =164 10. 2x – 3 – 23 = 56 9

Lesson 4 Solving Problems Involving Expressions with ExponentsScientific Notation: It is not uncommon in scientific applications of algebra to find yourself workingwith very large or very small numbers. Even in the time of Archimedes (287 – 212 B.C.),The study of such numbers was not unusual. Archimedes estimated the universe was23,000,000,000,000,000 m in diameter, which is the approximate distance light travelsin 2 1 years. 2 In scientific notation, he estimated the diameter of the universe would be 2.3 x1016 m. In general, you can define scientific notation as follows:Scientific Notation A number is in scientific notation if it is expressed as aproduct of two factors, one factor being less than 10 andgreater than or equal to 1and the other a power of 10expressed in exponential form.Examples:Write each of the following numbers in scientific notation.1. 120,000 = 1.2 x 105 The exponent of 10 is 5.2. 88,000,000 = 8.8 x 107 The exponent of 10 is 7.3. 420,000,000 = 4.2 x 1084. 3,000,000,000 = 3 x 109 Note the pattern in writing a number in scientific notation. The decimal point wasmoved to the left so that the multiplier will be a number between 1 and 10. The numberof places will be the exponent of 10. If you move to the left, the exponent is positive. However, if the decimal point is to be moved to the right, the exponent will benegative. 10

5. 0.0006 = 6 x 10-4 6. 0.0000000079 = 7.9 x 10-9Note: To convert back to standard or decimal form, the process is simply reversed.Examples:Write each of the following scientific notation in standard form.1. 3.1 x 105 The exponent is positive 5 so you move the 3. 1 0 0 0 0 decimal point 5 places to the right.Therefore, 3.1 x 105 = 310,000 in standard form2. 8.5 x 109 The exponent is positive 9 so you move the 8. 5 0 0 0 0 0 0 0 0 decimal point 9 places to the right. Add zeros if needed.Therefore, 8.5 x 109 = 8,500,000,000 in standard form3. 6.3 x 10-3 The exponent is negative 3 so you move the 0. 0 0 6 3 decimal point to the left. Add zeros if necessary.Therefore, 6.3 x 10-3 = 0.0063 in standard form4. 2.8 x 10-6 The exponent is negative 6 so you move the 0. 0 0 0 0 0 2 8 decimal point to the left. Add zeros if necessary.Therefore, 2.8 x 10-6 = 0.0000028 in standard form Scientific notation is useful in expressing large and small numbers specially whenyou are solving statement problems.Examples: 11

Solve the following problems.1. Light travels at a speed of 3.05 X 108 meters per second (m/s). There are approximately 3.15 x 107 in a year. How far does light travel in a year?Solution: Multiply the distance traveled in 1 sec by the number of seconds in ayear.This gives (3.05 x 108)(3.15 x 107) = (3.05 x 3.15)(108x107) = 9.6075 x1015You multiply the coefficients and add the exponents.2. The distance from earth to the star Spica ( in Virgo) is 2.2 x 1018 m. How many light-years is Spica from earth?Solution:2.2x1018 = 2.2x1018−16 Divide the distance (in meters) by 1016 the number of meters in 1 light- = 2.2 x 102 year. = 220 light yearsTry this outA. Which numbers are in scientific notation?1. 0.92 x 1022. 1.62 x 10-53. 10.63 x 1034. 8.26 x 1045. 0.34 x 10-1Each of the standard numerals is given in scientific notation. The scientificnotation may not be correct. Identify the numbers that are correct.6. 47,000 = 4.7 x 1047. 31,400 = 3.14 x 104 12

8. 35,604,000 = 3.56 x 107 9. 4,620,000 = 4.62 x 107 10. 179,990,000 = 1.8 x 108B. Write in scientific notation. 1. 120,000 2. 0.00078 3. 0.165 4. 1235 5. 678,435,848 6. 3,784,600 7. 0.0000000543 8. 0.00000077 9. 25.3 x 105 10. 39640000 x 10-3C. Write the following in standard form. 1. The rocket is 3.8 x 105 km above the earth. 2. The satellite travels 4 x 10 kilometers per minute. 3. Sound travels 9 x 104 meters per minute in water. 4. Water has a weight of 1 x 1011 kilograms per cubic kilometers. 5. There are about 2.2 x 108 molecules in an atom.D. Solve the following problems. 6. The diameter of a hydrogen atom is 0.00000001 cm. Express the diameter in scientific notation. 7. The mass of the earth is about 5795 000 000 000 000 000 000 kg. Express the mass in scientific notation. 8. The farthest object that can be seen with the unaided eye is the Andromeda galaxy. This galaxy is 2.3 x 1022 km from the earth. What is this distance in light-years? 13

9. A light year the distance travels in one year is equal to 9.44 x 1012 km. If the Polaris is about 64 000 000 000 km from the earth, how long will it take the light from this star to reach the earth? 10. The speed of radio waves is 297 600 km per second. How much time is needed for the radio impulse to travel from a station to a radio if the distance between them is 1280 km? Let’s summarize1. For any real numbers a, and integers m and n, the following laws of exponentapplies to expressions having the same base: a. Multiplication: am · an = am + n b. Powers of Power (am)n = amn. c. Division: am ÷ an = am - n2. Zero Exponents: For any real number a where a ≠ 0, a0 = 13. Negative Exponents: For any nonzero real number a and whole number n, a −n = 1 and a-n is the anmultiplicative inverse/ reciprocal of an.4. Exponential Equations are equations such that the unknown is an exponent.5. Scientific Notation A number is in scientific notation if it is expressed as a product of two factors,one factor being less than 10 and greater than or equal to 1and the other a power of 10expressed in exponential form. 14

What have you learnedA. Simplify the following expressions.1. 850 6. 4w-22. –8t03. (m4n2)0 7. (4w)-2 8. c −54. 16 + b0 a0 d −75. (-2)3(3)0 9. 1 x −4 (a −2 )3 b 4 10. b −2B. Simplify the following:1. (5)4 6. (4a5)(5a3)2. 52 · 53. (–4t)3 7. (5w-2)(4w)-24. (m4n2)3 8. 45x7y4 ÷ 5x3y35. (-5)3(3)2 9. 20c−5 5c−7 ( x−2 )3 y 4 x−2 y 10.C. Solve for x. 6. 6x – 3 – 52 = 191 1. 83x + 2 = 820 7. 162x + 1 = 4x - 3 2. 62x – 4 = 616 8. 82x = 16x - 1 3. 35x = 243 2x+1 4. 272x = 813 9. 27 3 =81 5. 252x = 625 1 +3x 10. 25 2 = 125D. Write in scientific notation. 1. 170,000,000 2. 8,200,000,000 3. 0.00000035 15

4. 0.000000000065E. Write the following in standard form 5. 4.0073 x 108 6. 2.791 x 10-4 7. 4.756 x 10-5F. Solve the following problems. 8. The mass of the sun is approximately 1.98 x 1030 kg. If this were written in standard or decimal form, how many zeros would follow the digit 8? 9. Megres, the nearest of the big Dipper stars is 6.6 x 1017 m from earth. Approximately how long does it take light traveling at 1016 m/year, to travel from Megres to earth? 10. The number of liters of water on earth is 15, 500 followed by 19 zeros. Write this number in scientific notation. Then use the number of liters of water on earth to find out how much water is available for each person on earth. The population is 5.3 billion. 16

ANSWER KEY 6. x8 7. 10a7b3How much do you know 8. 6a2b3c2A. 1. 81 9. 4x7y5 10. 4x4y4 2. 625 6. -12 3. 32768 7. 9 4. 8 8. 1 5. 9 9. 1B. 1. 1 10. 3 2. 1 6. y 4 3. 5 4. 1 x4 5. 7 7. 1C. 1. 1 x4 y4 3 8. 64 2. 1 9. y 3 100 x3 3. 1 10. b6 512 a6 4. 1 17 5. y 3 x3D. 1. x = 3 2. x = 3 3. x = 3

4. x = 25. x = 66. 1.8468 x 1077. 18,450,000,000,000,000,0008. 3 x 105 km/sec9. 6.02 x 102510. 6.24 x 1021Try this outLesson 1 6. 64A. 1. 10242. 125 7. 4199043. 93312 8. 12964. 100 000 9. 5125. 7776 10. 12 500B. 1. 4096 6. b302. 49 7. 16b63. 256 8. 125x15y94. 64 9. 16c10d6e25. 25 10. 9a6b4c2C. 1. 100 6. r2. 81 7. 7a153. 625 8. 9x3y244. 7 9. 3x0y35. 256 10. 6a4b2c0 18

Lesson 2 6. yA. 1. 1 7. 1 8. 1 2. 1 9. 1 3. 8 10. -5 4. –7 5. p 6. 10 x5B. 1. 1 a10 7. 1 16 y 4 2. 1 16 8. − 5 t2 3. - 1 16 9. x4 10. 2a 2 4. 4 25 3 5. 3 6. x = 2 w4 7. x = 3 8. x = 3Lesson 3 9. x = 3A. 1. x = 4 10. x = -5 2. x = 1 6. x = 2 3 7. x = 11 19 3. x = -3 4. x = 4 5. x = 2B. 1. x = 2 2. x = -1

3. x = 7 8. x = 1 4. x = 10 9. x = -2 5. x = 4 10 x = 9Lesson 4 6. CorrectA. 1. No 7. Correct 2. Yes 8. Wrong 3. No 9. Wrong 4. Yes 10. Wrong 5. No 20B. 1. 1.2 x 10 5 2. 7.8 x 10 -4 3. 1.65 x 10-1 4. 1.235 x 10 3 5. 6.78435848 x 108 6. 3.7846 x 106 7. 5.43 x 10-8 8. 7.7 x 10-7 9. 2.53 x 106 10. 3.964 x 104C. 1. 380,000 km 2. 40 km. 3. 90,000 m 4. 100,000,000,000 kg

5. 220,000,000 molecules 6. 4D. 6. 1 x 10-8 w2 7. 5.795 x 1021 7. 1 8. 2,300,000 16w2 9. 2.47 days 10. 4.3011 x 10-3 8. d 7What have you learned c5A. 1. 1 9. x4 2. –8 10. b6 3. 1 a6 4. 17 6. 20a8 7. 5 5. –8B. 1. 625 16w4 8. 9x4y 2. 125 9. 4c2 3. -64t3 10. y3 4. m12n6 x4 5. 1125C. 1. x = 6 2. x = 10 3. x = 1 21

4. x = 2 5. x = 1 6. x = 6 7. x = − 5 3 8. x = -2 9. x = 1 2 10. x = 1 3D. 1. 1.7 x 108 2. 8.2 x 109 3. 3.5 x 10-7 4. 6.5 x 10-11E. 5. 400,730,000 6. 0.0002791 7. 0.00004756F. 8. 28 9. 66 years 10. 1.55 x 1023 and 2.9 x 1013 22

Module 2 Radical Expressions What this module is about This module is about radical expressions and rational exponents. Some lessonsinclude transforming radical expressions into exponential form and vice versa. You willalso develop skills in simplifying radical expressions and expressions with rationalexponents. Several activities are provided for you in this module. What you are expected to learn This module is designed for you to: 1. define rational exponents. 2. write expressions in radical or exponential form. 3. simplify expressions involving rational exponents. How much do you knowA. Express the following in radical form.1.(9)12 27 1 125 6.  3  2.(81) 1 4 ( )1 2 7. 102 23.x 3 4 1 24 8.  2   25 1 814.  2   9. − (144 )1 25.(3x )1 ( )1 3 10. 4b10 2

B. Express the following in exponential form.1. 64 6. 3 10m2 7. 4 952. 3 8m2 8. 163 9. 4 24b43. ab 10. x 2 y 2( )4. 4 24 5 ( )35. 5xyC. Change the indicated roots into radicals and evaluate.1.(16)12 6.(16)432. − (121)12 7.(25)323.(169)12 8.(− 32) 3 54.(64 )1 9.(36) 5 3 25.(8 )2 10.(64)65 3 What you will do Lesson 1 Write Expressions with Rational Exponents as Radical Expressions and Vice Versa In the radical expressions n x we shall call n the index. The exponent of x in this 1expression is 1. n x can be transformed as the rational expression x n and vice versa. 1Notice that in the rational expression x n , the denominator n of the exponent is the indexand the numerator 1 is the index of the radical expression n x . You can use this knowledge to write expressions involving rational exponents asradicals and vice versa.Writing Rational Expressions in Radical Form:

1 If n is a positive integer, xn = n x .Examples:Write each rational expression in radical form. 1 The denominator is the index. If the index is 2, there is no need of writing.1) 252 = 25 1 The denominator is 3 and so the index is 3.2) 83 = 3 8 1 The denominator is 4 and so the index is 4.3) 814 = 4 81 If m is any rational number where n ≠ m = n xm = (n x )m . n 0, x nExamples:Write each rational expression in radical form. 3 m1) a5 = 5 a3 Here we use a n = n am , which is generally the preferred form in this situation.2) (mn )3 = 4 (mn)3 = 4 m3n3 The exponent applies to mn because of 4 the parenthesis. 5 Note that the exponent applies only to the variable y.3) 2y 6 = 26 y5 Now the exponent applies to4) (24 )5 =6 (2y)5 = 6 32y5 2y because of the parenthesis. 6Examples:Write each rational expression in radical form.1) (ab)32 3) (3x )3 4 3 22) 3x 4Solutions: 4) x7

1) (ab )2 = 3 (ab)2 = 3 a2b2 3 32) 3x 4 = 34 x33) (3x )3 = 4 (3x)3 = 4 27x3 4 ( )2 24) x7 = 7 x2 or 7 xWriting Radical Expressions in Exponential Form: 1 In writing radical expressions in exponential form such as n x = x n , where n is apositive integer, the index is the denominator while the exponent is the numerator.Examples:Write each radical expressions in exponential form.1) 5 x 3) 5 2a2) 4 x3 ( )3 4) 5 ySolutions: 1 The index is 5 and the exponent is 1.1) 5 x = x5 The index is 4 and the exponent is 3. 3 The index is 5 and the exponent is 1.2) 4 x3 = x 4 The index is 2 and the exponent is 3. Five is not included in the parenthesis so the fractional3) 5 2a = (2a)51 exponent is only for y. () 3 34) 5 y = 5y 2Try this outA. Write each radical expressions in radical form. 3 6) (2y )3 41) a4 5 32) m 6 7) 9 2

2 33) 2x3 8) 2 y 4 2 34) 3x5 9) 16 45) (3x )2 2 5 10) 8 3B. Write in exponential form1. 7a 6) 5 32r s10 152) 3 27p6 7) 64x23) 4 81x 8 y16 8) 5 32y54) 25w 45) 3 8m6n9 9) 121 10) 3 64C. Math Integration Where Is the Temple of Artemis? The temple of Artemis is one of the seven wonders of the world. It was builtmostly of marble around 550 B.C. in honor of a Greek goddess, Artemis. In what country can this temple be found? To answer, simplify the following radical expressions. Cross out each box thatcontains an answer. The remaining boxes will spell out the answer to the question.1) 3 8 = ___________2) 4 163 = ___________ = ___________ ( )3 = ___________ = ___________3) - 44) 455) 4 25

6)  3 27 2 = ___________ 8( )7) 4 81 3 = ___________( )8) 6 64 4 = ___________ GT EUGRE 2 9 27 1 -8 12 8 YPKECTY 2 32 4 1 16 9 3 5 24 4Answer: _____ _____ _____ _____ _____ _____Source: Math Journal Volume X – Number 3 SY 2002-2003 Lesson 2 Simplifying Expressions Involving Rational Exponents We shall use previous knowledge of transforming rational expressions toradical to simplify rational expressions.Examples:Express the given expression in radical form and then simplify. 11. a = 42Solution: (a)2 =  4 1 2 Square both sides 2 In the equation a2 = 4, you can see that a is the a2 = 4 number whose square is 4; that is; a is the principal square root of 4. a= 4 2 is the principal square root of 4

a=2 12) 25 2 = 25 = 5 1 27 1/3 is the cube root of 27.3) 27 3 = 3 27 = 3 − 36 is not real number 32 1/5 is the fifth root 1 of 32.4) − 362 = − 36 = 65) (− 36)12 = − 36 16) 32 5 = 5 32 = 2 m an =( )For any real number a and positive integers m and n with n >1, n a m = n am .The two radical forms for a ( )mn n a m and n a m  are equivalent, and the choice of  which form to use generally depends on whether we are evaluating numericalexpressions or rewriting expressions containing variables in radical form.Examples: Simplify expressions with rational exponents.( )3 3 Write the rational expression as radical expression.1) 92 = The denominator 2, is the index. The numerator 3, 9 is the exponent of 9 or the radical expression 9 .= (3)3 Simplify= 27Here, you take 9 , then cube the result. This will give you the answer 27.

2) 3 =  4 16 3 Express the given expression in radical form. 81 The denominator 4 is the index. 16 4 The numerator 3 is the exponent of the radical  81 expression. The fourth root of 16 is 2. =  2 3 The fourth root of 81 is 3. 3 Multiply 2 by itself three times or you take the =8 3 27 cube of 2 .  2  2  2  = 8( )3)(− 8 )2 = 3 −8 2 3  3  3  3  27 3 Write in radical form. Get the cube root of –8, it = (− 2)2 is –2. Then, you square it. =4Note that in example 1, you could also have evaluated the expression as 3 9 2 = 93 = 729 = 27. () m This shows why we use n a m for a n when evaluating numerical expressions.The numbers involved will be smaller and easier to work withTry this out 1A. Use the definition of a n to evaluate each expression. 1 6) (- 64)31 1) 362 1 1 7) 814 2) 1002 1 1 8) − 32 5 3) 252 1 4) (64)21 9)  4 2 9 1 1 5) 273 10)  27 3 8