| 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.
Let's substitute V=64 into the formula and solve for s.
Solve for s
3LHS=3RHS364=3s3Calculate root4=sRearrange equation
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.
Let's substitute V=216 into the formula and solve for s.
Solve for s
3LHS=3RHS3216=3s3Calculate root6=sRearrange equation
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.
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.
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.
For this rectangle ℓ=3729 and w=41/2.
Solve for A
Calculate rootA=9⋅41/2Calculate powerA=9⋅2Multiply
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.
Now lets substitute V=275 and solve for s.
Solve for s
3LHS=3RHS3275=3s3Simplify radical and power275/3=s3/3Calculate power243=sRearrange equation
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=327527=333s3=3(33)5Rewrite (33)5 as (35)33s3=3(35)3Calculate 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.
Solve for V
Calculate powerV≈63.141(464.76)Calculate rootV≈6(1.77)1(464.76)MultiplyV≈10.631(464.76)Calculate quotientV≈(0.094)(464.76)Multiply
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.
Let's substitute A=x.
Solve for s
LHS=RHSx=s2Calculate root±x=sRearrange equations=±xRewrite x as 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.
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.
Solve for r
Calculate quotientr=(1.375)1/6−1Calculate powerr≈1.0545−1Subtract terms
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.
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.
Solve for r
Calculate quotientr≈(2.4178)1/10−1Calculate powerr≈1.0923−1Subtract terms
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.
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.
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
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.
Solve for ℓ
LHS/7.66=RHS/7.662.61≈ℓ3LHS31=RHS312.611/3≈(ell3)1/3Calculate power1.37≈ℓRearrange equation
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 || |