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FEATURING: Joe, the boy with mutated cells! by: Mahika Gupta, Mihir Kekkar,
jake jiang (Period 3, Group 8) Appendix Chapter 1.5 ……………………………………………………………… pages 2­3 Chapter 1.6 ……………………………………………………………… page 4 Chapter 1 Practice Problems ……………………………………………. page 5 Chapter 4.7 ……………………………………………………………… pages 6­7 Chapter 4 Practice Problems ……………………………………………. page 8 Chapter 6.5 ……………………………………………………………… page 9 Chapter 6.6 ……………………………………………………………… pages 10­11 Chapter 6 Practice Problems ……………………………………………. page 12 Science Problem ………………………………………………………… page 13­14 Practice Problem Solutions ……………………………………………… page 15­16 1.5: Equations I​
n mathematics, an ​
equation​
is an equality containing one or more variables. Solving the equation​
consists of determining which values of the variables make the equality true. In this situation, variables are also known as unknowns and the values which satisfy the equality are known as solutions. Example 1 Solve 3x­5=16 Solution: The technique is to transform the equation to x=constant by getting rid of the unwanted numbers around the x in the left member. 3x­5=16 Write the given equation 3x=21 Add 5 to each member x=7 Multiply each member by 1/3 This equation is equivalent to the original one, meaning it has the same solution set. By inspection, you can write S={7}. Here’s another example: Solve (3x­4)(x+5)=0 Solution: This equation contains a product which equals zero. The only way to product of two real numbers can equal zero is for one of the factors to equal zero. This fact is expressed in the converse of Multiplication Property of Zero, which can be proved from the axioms as in Section 1­7. Because of this fact, you can transform the equation to 3x­4=0 or x+5=0 Further transformations give 3x=4 or x=­5
Adding 4 or subtracting 5. x=4/3 or x=­5 Dividing by 3. S={4/3,­5} Solving the transformed equation. Example 1 ​
Solve . Solution In this equation the basic problem is the square root. If that weren’t there we could do the problem. The whole process that we’re going to go through here is set up to eliminate the square root. However, as we will see, the steps that we’re going to take can actually cause problems for us. So, let’s see how this all works. Let’s notice that if we just square both sides we can make the square root go away. Let’s do that and see what happens. 1.6:​
Inequalities If the “=” sign in an equation is replaced by one of the order signs, <,>, ≤, ​
≥, the resulting sentence is called an inequality. For example, 3x ­ 5 < 16 The solution set of an inequality contains all values of the variable that make the sentence true. Since the solution set usually contains an infinite number of solutions, it is customary to draw a graph rather than write the set. Example: Solve: x + 3 < 7 Answer: If we subtract 3 from both sides, we get: x + 3 ­ 3 < 7 ­ 3 x < 4 And that is our solution: x < 4 In other words, x can be any value less than 4. What did we do? We went from this: x + 3 < 7 To this: x < 3 4.7: Solution of Systems by Augmented Matrices Augmented Matrices:​
Matrices that are used to solve a system of linear equations by separating x coefficients from the y term. Example​
: 5x + 4y = ­30 x + 3y = ­6 Objective: Simplify Matrix into RREF (Reduced Row Echelon Form) *​
The objective is to transform the matrix into this form: R2
​
2.75 → R1 ­ 5R2 → R2
R1
5 & −4 →
R1 + R2 → R1 ​
x = ­6 ; y = 0 Step 1:​
Objective is to make a Zero ­ Since the matrix is just two equations, all the properties of those equations apply to the matrix ­ By multiplying the second row of the matrix by 5, you have two fives as the coefficient of the first terms in the equation ­ To get rid of them, you can subtract them to get a 0 for the first term in the second row Step 2: ​
Objective is to make another Zero ­ In order to make another zero, the two coefficients of the second terms have to be the same or opposites ­ To make this happen you can divide the second equation by 2.75 ­ Since the two other terms are 0, dividing them won’t change anything Step 3: ​
Objective is to make another Zero ­ Now that the two coefficients of the second terms are opposites, you can add both equations and plug in the result to row 1 to get a zero for the second term Step 4: ​
Objective is to make Ones ­ Now that all the zeroes are taken care of, the rest of the coefficients on the left side have to be one so that the equations read x + 0 = ? and y + 0 = ?. ­ To do that, you just divide each row by the last coefficient remaining ­ In the first row, that is 5 so you divide that row by 5 ­ In the second row, that is ­4 so you divide that row by ­4. This completes the matrix. In equation form, this reads 1x + 0 = ­6, which is 1x = 6, and 1y+0=0, which is y=0. There is another harder example problem below: 3x + 8y = 54 4x + 5y = 38 R1 ­ 34 R2 → R2 R1
3 →
R2
​
4.25 →
R1 ­ 8R2 → R1
​
6.5 Powers and Radicals without Calculators Objective: Be able to calculate expressions that contain fractional exponents or radicals quickly in your head and only use a calculator to check your answer. 2/3
Example 1:​
Evaluate 27​
Solution: 3​
2​
y/z​
= (​
√27)​
: When dealing with fractional exponents, you can rewrite x​
as z​ y​ z​ y​
(​
√x)​
or ​
√(x​
). For this expression, using the denominator first is more efficient because 27 is a perfect cube. However, some expressions are better to be raised to a power before rooted, and it is very useful to tell which is better in an expression. 2​
= 3​
:
Evaluate the first step, based on whether you decided to use the exponent’s denominator or numerator first. In this expression, the denominator was used, 3​
and ​
√27 is 3. 2​
= 9:
3​
is 9 Objective: Use this method in reverse to turn a radical into a fractional exponent, which can also be used to simplify expressions which normally cannot be rooted rationally. 2​
3​
Example 2:​
Simplify ​
√27 ÷ ​
√243 Solution: 1/2​
1/3​
=27​
* 243​
: Simplify the radical back into exponential form. The square root of 27 is equal to raising it to the ½ power. The cube root of 243 is the same as raising it to the ⅓ power. 3​1/2​
5​⅓​
=(3​
)​ * (3​
)​
:Reduce the terms to the lowest possible common base. 27 can be 3​
5​
reduced to 3​
and 243 is 3​
. 3/2​
5/3​
=3​ * 3​:
Multiply the exponent outside of the parenthesis with the one inside the parenthesis for each term. 19/6​
=3​
:
The bases are the same, so you can add the exponents. 6.6: Scientific Notation Why is it used? Scientific Notation is used to be able to multiply large numbers quickly and easily using decimals and powers of 10. 15
2,364,597,285,765,625​
in scientific notation is ​
2.364597285765625 * 10​
However, the beauty of scientific notation is that it is normally done with rounding. So, in this case, the number would be: 15 2.36 * 10​
Numbers written in scientific notation usually contain 3 significant figures: one before the decimal and two after. When plugging in especially large numbers into a calculator, the calculator will usually express the number in scientific notation. It usually looks like this: 2.3645973 15 15 This means the exact same thing: ​
2.3645793 * 10​
A number written in scientific notation has two different components: the ​
mantissa ​
and the characteristic​
. characteristic A number is ​
only​
in scientific notation if it is written in this form: ​
mantissa * 10​
where the ​
mantissa​
is a number between 1 and 10 (including 1 but not 10) and the characteristic ​
is an integer that tells how many places the decimal point has been moved. Example 1: ​
Write ​
0.00000076848​
in scientific notation a. The mantissa has to be between 1 and 10, which means it has to be a one digit number i.
7.6848 b. But, this number isn’t the same as the number given in the problem. ­7
i.
7.6848 * 10​
­7​
c. 10​
is equivalent to 0.0000001, and that multiplied to 7.6848 is 0.00000076848 d. You can round the answer to the nearest hundredth ­7 i.
7.68 * 10​
*If you make one factor smaller by moving the decimal point, you must make the power of 10 larger (by increasing its exponent), and vice versa, so that the number you get for the answer equals the number you started with. 23​
­20​
Example 2: ​
Multiply ​
(6 * 10​
)(4 * 10​
)​
in your head. Write the answer in scientific notation. a. Using the commutative property of multiplication, you can move around the numbers so that you are multiplying the mantissas and the exponents. 23​
­20
i.
6 * 4 * 10​
* 10​
b. You can use the exponential property of multiplication of exponents to add the exponents together. 3
i.
24 * 10​
c. Now, all that is left is to simplify this into scientific notation. The first thing to do is make the mantissa less than 10. 3
i.
2.4 * 10​
d. By dividing 24 by ten, we made the mantissa 2.4 which fits the requirements for scientific notation. But, since we made the mantissa smaller, we have to make the characteristic bigger. We divided the mantissa by ten, so we increase the characteristic by one. 4 i.
2.4 * 10​
Chapter 1 Practice Problems: 1. 7(2x­3) ­ 4(x+5) = 8(x­1) + 3 2. 7 = √x + 8 3. 5​
x​
+ 7 < 3(​
x​
+ 1) 4. 3(​
x​
– 2) + 4 ​
>​
2(2​
x​
– 3) Chapter 4 Practice Problems: 1. ­6x ­ 8y = ­2 ­3x + 2y = ­49 2. 8x ­ 2y = 40 ­4x ­9y = ­140 Chapter 6 Practice Problems: 3/4 1. 81​
5/2 2. 4​
2/3 3. 343​
4.​
Write 6,453,874,000 in scientific notation. Round to 3 significant figures. 32​
23​
5. Multiply (5.6 * 10​
)(4.3 * 10​
). Write answer in scientific notation. Injections Against Infections (Challenge Problem) Joe has gone into the doctor’s office to get checked, because he recently caught the flu. After taking some scans of Joe’s body, along with a blood test, the doctor comes to the conclusion that ­2​
Joe has 4.7 * 10​
mL of his bloodstream consisting of the malignant pathogen. His doctor decides to try out a new medication. He prescribes Joe with an injection, which has a volume of ­3​
8*10​
mL per dose. Joe needs to take twice as much of the injection as the amount of volume of his blood that has the flu pathogen. At least how many whole doses of the injection does Joe need to fully meet his needs? Solution to Challenge Problem 1. Understanding the information and the question: Joe needs to get at ­2​
­3​
least 2(4.7 * 10​
) mL of his injection in increments of 8*10​
mL. 2. Setting up the equation: This means the exact amount of doses he ­2​
­3​
needs is [2(4.7 * 10​
)] ÷ (8 * 10​
). 3. Arithmetic: Joe will need exactly 11.75 doses. 4. Meeting the question’s requirements: The question asks for the minimum whole doses needed, so the exact answer will have to be rounded up to the nearest integer value. 5. Final Answer: Joe needs at least 12 whole doses of the injection. This question shows how scientific notation is important in real life events. It helps us quickly compute large numbers without much work and is great for estimation. Answer Key: 1.5: 1.6: 1. 5​
x ​
+ 7 < 3(​
x​
+ 1) 1. 7(2x­3) ­ 4(x+5) = 8(x­1) + 3 Answer: 14x­21­4x­20=8x­8+3 Answer: 2​
2 7​
= ( √x + 8 )​
49 = x + 8 x = 41 Answer: 5​
x​
+ 7 < 3(​
x​
+ 1) 5​
x​
+ 7 < 3​
x​
+ 3 2​
x​
+ 7 < 3 2​
x​
< –4 x​
< –2 2. 3(​
x​
– 2) + 4 ​
>​
2(2​
x​
– 3) Answer: 3(​
x​
– 2) + 4 ​
>​
2(2​
x​
– 3) 3​
x​
– 6 + 4 ​
>​
4​
x​
– 6 3​
x​
– 2 ​
>​
4​
x​
– 6 –2 ​
>​
​
x​
– 6 4 ​
>​
​
x x​
​
<​
4 4.7: 6.5: 1. ­6x ­ 8y = ­2 ­3x + 2y = ­49 3/4 1. 81​
Answer: √4 81 = 3 3 ​
3​
= 27 27 *You can easily solve this in your head by memorizing powers of 3. 5/2 2. 4​
Answer: √4 = 2 5​
2​
= 32 32 *You can easily solve this in your head by memorizing powers of 2. 10x­41=8x­5 ­8x ­8x 2x­41=­5 +41 +41 2x=36 x=18 2. 7 = √x + 8 Answer: R1
−2 → R1 + R2 → R2→ R2
6
→ R1­4R2→R1 R1
11
→ x = 11; y = 8 2. 8x ­ 2y = 40 ­4x ­9y = ­140 2/3 3. 343​
3
√343 = 7 2​
7​
= 49 49 *You can do this easily in your head by memorizing small powers of 7 R1 + 2R2 → R2 R2
R1
2 & −20 → R1
4 → x = 8; y = 12 R1 + R2 → R1 6.6: 1. Write 6,453,874,000 in scientific notation. Round to 3 significant figures. Answer: 9 6.45 * 10​
32​
23​
2.​
​
Multiply (5.6 * 10​
)(4.3 * 10​
). Write answer in scientific notation rounded to 3 significant figures. Answer: 32​
23
5.6 * 4.3 * 10​
* 10​
55 24.08 * 10​
56 2.41 * 10​