Radicals and Rational Exponents

Download for free
Find the solutions in the app
Android iOS
Exercises marked with requires Mathleaks premium to view it's solution in the app. Download Mathleaks app on Google Play or iTunes AppStore.
Sections
Communicate Your Answer
Exercise name Free?
Communicate Your Answer 3
Communicate Your Answer 4
Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Monitoring Progress 8
Exercises
Exercise name Free?
Exercises 1
Exercises 2
Exercises 3
Exercises 4
Exercises 5
Exercises 6
Exercises 7
Exercises 8
Exercises 9
Exercises 10
Exercises 11 When given the volume and asked to find the length of a side, it's best to remind yourself of the formula for the cube's volume. Vcube​=s3​ Let's substitute V=64 into the formula and solve for s. V=s3V=6464=s3 Solve for s 3LHS​=3RHS​364​=3s3​Calculate root4=sRearrange equation s=4 The side of a cube with volume 64 in3 is 4 in.
Exercises 12 When given the volume and asked to find the length of a side, it's best to remind yourself of the formula for the cube's volume. Vcube​=s3​ Let's substitute V=216 into the formula and solve for s. V=s3V=216216=s3 Solve for s 3LHS​=3RHS​3216​=3s3​Calculate root6=sRearrange equation s=6 The side of a cube with volume 216 in3 is 6 in.
Exercises 13
Exercises 14
Exercises 15
Exercises 16
Exercises 17
Exercises 18
Exercises 19
Exercises 20
Exercises 21
Exercises 22
Exercises 23
Exercises 24
Exercises 25
Exercises 26
Exercises 27
Exercises 28
Exercises 29
Exercises 30 When describing errors, it's important to justify each step to see if you can identify the error. Let's look at the first step. (-81)3/4=[(-81)1/4]3​ In this step we can see that the rule of (am)n=am⋅n was applied where m=1/4 and n=3. This step follows the rules for exponents, lets look at the next step. ​==​[(-81)1/4]3(-3)3​ In this step, there is an error. We cannot take an even root of a negative number because answers to even exponents are always positive. Since x4=-81 has no real solution, we cannot calculate the given expression's answer.
Exercises 31
Exercises 32
Exercises 33
Exercises 34
Exercises 35 When asked to find the area, we need to identify the geometric figure, to determine which formula to use. In this case we are looking at a rectangle. ARectangle​=ℓ⋅w​ For this rectangle ℓ=3729​ and w=41/2. A=ℓ⋅wℓ=3729​, w=41/2A=3729​⋅41/2 Solve for A Calculate rootA=9⋅41/2Calculate powerA=9⋅2Multiply A=18 There for the area of the bake sale sign is 18 ft2.
Exercises 36 When a number is written in power form, we tend to want to compute it. In this exercise, 275 is a very large number, let's leave it like it is and see what happens as we solve. First, let's recall the formula for a cube's volume. Vcube​=s3​ Now lets substitute V=275 and solve for s. V=s3V=275275=s3 Solve for s 3LHS​=3RHS​3275​=3s3​Simplify radical and power275/3=s3/3Calculate power243=sRearrange equation s=243 Therefore, if a cube has a volume of 275 mm3, then the length of its side is 243 mm. Alternative solution Mental Math Another way to think about 275 is to rewrite 27 as a perfect cube, then take the cubed root. These steps can be done mentally, so they can eliminate the need for a calculator. 3s3​=3275​27=333s3​=3(33)5​Rewrite (33)5 as (35)33s3​=3(35)3​Calculate roots=35Calculate powers=243
Exercises 37 In this exercise, we are given a formula and a picture with values. We need to substitute the values from the picture into the formula and solve. Given the information, we can see that V=5, h=4, and they want us to use π=3.14. r=(πh3V​)1/2Substitute valuesr=(3.14(4)3(5)​)1/2Multiplyr=(12.5615​)1/2Calculate quotientr≈(1.194)1/2Calculate powerr≈1.093 From this we can say that the radius of the cup to the nearest inch is 1in.
Exercises 38 In this exercise, we have a formula and a value for surface area. Let's substitute the value, S=60 into the formula and solve. V=6π​1​S3/2V=60, π=3.14V=63.14​1​(60)3/2 Solve for V Calculate powerV≈63.14​1​(464.76)Calculate rootV≈6(1.77)1​(464.76)MultiplyV≈10.631​(464.76)Calculate quotientV≈(0.094)(464.76)Multiply V≈43.71 From this we can approximate the volume of the sphere to be 44m3.
Exercises 39
Exercises 40 For this exercise we need to write an expression for the the side length of a square with area x in2. Let's start with the usual formula for area then substitute our value for area. As​quare=s2​ Let's substitute A=x. A=s2A=xx=s2 Solve for s LHS​=RHS​x​=s2​Calculate root±x​=sRearrange equations=±x​Rewrite x​ as x1/2 s=±x1/2 In this exercise, we are talking about the side length of a square, so we can ignore the negative solution and say s=x1/2 in., when the area is x in2.
Exercises 41 To calculate the annual inflation rate, we can use the formula. r=(PF​)1/n−1​ P is the initial value, in our exercise, P=800000. F is the new value, in this case, F=1100000. And n is the number of years, in our case, n=6. Let's substitute those values into our equation and solve. r=(PF​)1/n−1Substitute valuesr=(8000001100000​)1/6−1 Solve for r Calculate quotientr=(1.375)1/6−1Calculate powerr≈1.0545−1Subtract terms r≈0.0545 To convert that decimal to a percent, we need to multiply by 100 or simply move the decimal two places to the right, yielding r≈5.45%. For this exercise, we round that to the nearest tenth of a percent and we get a 5.5% rate of inflation over the 6 years.
Exercises 42 To calculate the annual inflation rate, we can use the formula. r=(PF​)1/n−1​ P is the initial value, in our exercise, P=1.46. F is the new value, in this case, F=3.53. And n is the number of years, in our case, n=10. Let's substitute those values into our equation and solve. r=(PF​)1/n−1Substitute valuesr=(1.463.53​)1/10−1 Solve for r Calculate quotientr≈(2.4178)1/10−1Calculate powerr≈1.0923−1Subtract terms r≈0.0923 To convert that decimal to a percent, we need to multiply by 100 or simply move the decimal two places to the right, yielding r≈9.23%. For this exercise, we round that to the nearest tenth of a percent and we get a 9.2% rate of inflation over the 10 years.
Exercises 43 When looking for solutions to equations that don't seem possible, consider numbers that always have the same solution for exponents. In this case 0 and 1. Let's look at why these work. x↕0​=↕=​x1/5↕01/5​ Zero to the power of any positive number is always zero, so this statement is true and one solution is x=0. Similarly, one to the power of any number is always one. x↕1​=↕=​x1/5↕11/5​ Therefore the second solution to this equation is x=1. To prove these are the only solutions is beyond the scope of this exercise.
Exercises 44 One way to check if our friend is correct is to check values of our own. We need to consider even and odd roots as well positive and negative arguments. Let's set up a table and compute different values for na​ and -na​.nana​-na​ 264264​=8-264​=-8 2-642-64​=D.N.E.-2-64​=D.N.E. 364364​=4-364​=-4 3-642-64​=-4-2-64​=-(-4)=4 In the first and third cases, it looks like our hypothetical friend is correct, but in the last case, we get a negative answer for na​ and positive one for -na​. In terms of logic, we only need one example to prove a theory false. We can generalize and say that our friend is correct if a is positive, but wrong when a is negative and n is odd.
Exercises 45
Exercises 46
Exercises 47
Exercises 48
Exercises 49 In this exercise we are given the formula for the volume of a dodecahedron, as well as a picture of one with a volume of 20 ft3. Let's substitute V=20 into the given formula and solve for ℓ, the edge length. V≈7.66ℓ3V=2020≈7.66ℓ3 Solve for ℓ LHS/7.66=RHS/7.662.61≈ℓ3LHS31​=RHS31​2.611/3≈(ell3)1/3Calculate power1.37≈ℓRearrange equation ℓ≈1.37 Therefore, the edge length of a dodecahedron with a volume of 20 ft3 is about 1.37 ft.
Exercises 50
Exercises 51
Exercises 52
Exercises 53
Exercises 54
Exercises 55
Exercises 56
Exercises 57
Exercises 58
Exercises 59
Exercises 60