Section 2.2- Applied Problems

Due to mediocre timing, mediocre technical issues and mediocre-ness in general, this blog post is a few weeks late, however, dont let that stop you from expanding your knowledge of applied problems!

First off, there are a few helpful steps to solving applied problems "easily". such guidelines are as follows.

1. Read the problem over several times carefully and identify the facts that are given, and the value that you hope to find.

2. Assign a letter to represent the unknown quantity.

3. If appropriate, draw and label a picture.

4. List all of the known facts and a relationships involving the unknown quantity. (relationships are simply equations with written words on one or both sides of the equal sign.)

5. Formulate and equation that states the relationship from step 4.

6. solve the equation from step 5.

7. check the solution to step 6 against the origional given problem.

Simple Example: Jack had 10 apples, and then he gave 4 of the apples to jane. how many apples remain?

1. we hope to find the number of apples

2. X will represent the number of remaining apples

3. no picture is necessary

4. 10 apples - 4 apples = X apples

5. 10 - 4 = X

6. 10 - 4 = 6

7. Jack had 10 apples, then he gave 4 to Jane. Jack was left with 6 apples

Obviously, this problem was so easy most of you probably lost brain cells just from reading it. and if you did have trouble with the math involved in this problem, please speak with your teacher immidialely. Next, we'll try a little more complicated problem involving perentages and discounts.

Example: Jimmy Dean went to the store to buy breakfast treats. he bought a sandwich for $2.59, a bagel for $3.19, and a pack of breakfast sausage. he paid a total of $7.02 after a 20% discount on all three items. how much was a pre-sale cost of the sausage?

1. we hope to find the cost of the breakfast sausages, and the total pre-sale cost of the food.

2. let X represent the price of the sausages, and Y represent the pre-sale food cost.

3. no picture is necessary.

4. $2.59 + $3.19 + price of sausage = total presale price

total presale price * o.20 = $7.02

5. $2.59 + $3.19 + X = Y

Y * 0.80 = $7.02

6. Y = $5.78 + X

0.80($5.78 + X) = $7.02

$4.62 + 0.80X = $7.02

0.80X = $2.4

X = $3.00

7. $3.00 + $3.19 + $2.59 = $8.78

$8.78 * 0.80 = $7.02

As you can see that problem was almost as easy as the first, just with more complicated numbers, but as long as you follow the steps for applied problems, you too can enjoy math! I hope this post helped you all with the homework you did weeks ago, and i wish you all a very mediocre day!

2.6 Inequalities

Hi Everyone!

Today in class we learned about inequalities, which are statements that express that two quantities or expressions are not equal by using the signs...

(less than)

(greater than)

(less than or equal to)

(greater than or equal to)

Solving inequalities is similar to solving equations except for a few differences:

When you divide by a negative number, you must switch the inequalities sign:

Ex.

You must also switch the sign if you take the reciprocal of both expressions:

Ex.

Also, when you're solving an inequality, you can't square both sides or take the square root of both sides.

After you solve an inequality you must put your answer into interval notation:

You use parenthesis to represent less than/greater than, and use brackets to express less than or equal to/greater than or equal to:

Ex.

Interval Notation:

Ex.

Interval Notation:

Ex.

Interval Notation: (- infinity, 5)U(8, infinity)

*U=Union

You can also show your answers to inequalities as graphs, like the ones below:

open dot: less than/greater than

closed dot: less than or equal to/greater than or equal to

So those are basically the only differences between inequalities and equations so...Let's solve some inequalities!

Ex.

divide each side of the inequality by 5

Interval Notation:

Ex.

One way to solve an inequality like this is to split it into two inequalities:

and

solve both inequalities by subtracting 7 from each side and you get...

and

Interval Notation: (-infinity, -6) U (-3, infinity)

A faster way to solve this inequality is by solving both inequalities at the same time:

subtract 7 from the whole inequality...

Interval Notation: (-infinity, -6)U(-3, infinity)

Ex.

First, multiply the whole inequality by 2

Then, divide the inequality by 3

Interval Notation:

Ex.

You know that the left side of the inequality has to be greater than zero, so in order for it to be greater than zero, the ratio must be positive. So, the denominator has to be positive to make the inequality true. Therefore....

subtract 4 from each side

Interval Notation: (-4, infinity)

Absolute Value Inequalities:

Solving absolute value inequalities is basically the same as solving absolute value equations.

Ex.

split the inequality into two inequalities: one with x-3 and another with -(x-3)

and

then, solve the inequalities in the same way you would solve a normal inequality...

and

Interval Notation: (1,5)

So, that's it. I hope this is helpful.

-Olivia

Chapter 2.5: Other Types of Equations

Solving an Equation Containing an Absolute Value:

Solving an Equation Using Grouping:

 Solving an Equation Containing Rational Exponents:

Solving an Equation Containing Radicals:

Solving an Equation of Quadratic Type:

2.4

Yet again,my computer wouldn't let me post when I pressed Publish Post, so instead of typing it up again on my sisters mac, I just made a website, so go here if you care anything about math......2.4 MATH BLOG of MAGNIFICENCE

By Josh (the coolest kid) Smith

2.3 Quadratic Equations

Alright everybody, so when working those quadratic equations there are several big things to remember.
Number 1: The Game
Second, is the Zero Factor Theorum.
This states that if p and q are algebraic expressions, then p=q if and only if p=0 or q=0.
Another important thing to know when solving them quadratic equations is completing the square.
This means that when:

We add to both sides to give us... the perfect square.
The example given to us by Mr. Wilhelm himself was:






Also, when solving quadratic equations, always keep in mind that if that means that . The key here is that it can be negative or positive.
Now last but not least is the Quadratic Formula.
If a is not zero, then...
x=(-b±√(b^2-4ac))/2a
This will give us two solutions of the equation. Be sure to check these people.
Now for your enjoyment...




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