## Day 166 – May 16

Algebra: Chapter 12/13 Review

The Benchmark #12/13 review packet is just like the test for Monday covering Chapters 12 and 13. Bring your questions and make sure that your notes are UP TO DATE! 200 points on the line!

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Algebra 1a: Chapter 9, Lesson 6, page 421 (day #2).

Graphing Systems of Linear Inequalities

We continue Chapter 9-5 techniques and solve 2 inequalities.

For the first equation, we can use any technique we learned from Chapter 7. Usually, the slope-intercept or `x` and `y` intercepts can be used to quickly define the line. Check a simple point like `(0, 0)` to see if that part of the ½ plane is true. If so, then shade that area.

Do the same for the other inequality and shade the appropriate ½ plane. The IMPORTANT PART is WHERE THE 2 INEQUALITIES OVERLAP THEIR SHADING, IS THE SOLUTION TO BOTH INEQUALITIES.

Remember, boundary lines of the form `<` or `>` are DASHED. The line is NOT part of the solution. Lines of the form `≤` or`≥` are solid because their line ARE part of the solution.

Again, the textbook is pretty good here but here are some more examples from purplemath.com too!

Two of tonight’s homework problems solved by MrE are here! Just click it.

## Day 165 – May 15

Algebra: Chapter 12 Review

A RELATION is a set of ordered pairs, a FUNCTION is a relation that assigns to each member of the DOMAIN (`x`) exactly one member of the RANGE (`y`).

• Remember the VERTICAL LINE TEST to determine a FUNCTION!
• A LINEAR FUNCTION is of the form `y = mx + b`
• A QUADRATIC FUNCTION can be written `y = ax^2 + bx + c`
• DIRECT VARIATION: `y = kx`
• INDIRECT VARIATION: ` y = k/x`
• JOINT VARIATION: `y = kxz` or `y = kx/z`

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Algebra 1a: Chapter 9, Lesson 6, page 421.

Graphing Systems of Linear Inequalities

We continue Chapter 9-5 techniques and solve 2 inequalities.

For the first equation, we can use any technique we learned from Chapter 7. Usually, the slope-intercept or `x` and `y` intercepts can be used to quickly define the line. Check a simple point like `(0, 0)` to see if that part of the ½ plane is true. If so, then shade that area.

Do the same for the other inequality and shade the appropriate ½ plane. The IMPORTANT PART is WHERE THE 2 INEQUALITIES OVERLAP THEIR SHADING, IS THE SOLUTION TO BOTH INEQUALITIES.

Remember, boundary lines of the form `<` or `>` are DASHED. The line is NOT part of the solution. Lines of the form `≤` or`≥` are solid because their line ARE part of the solution.

Again, the textbook is pretty good here but here are some more examples from purplemath.com too!

Two of tonight’s homework problems solved by MrE are here! Just click it.

## Day 164 – May 14

Algebra: Chapter 12, Lesson 7, page 565.

Joint and Combined Variation

An equation of the form `z = kxy`, where `k` is a constant, expresses joint variation. An equation of the form `z = (kx)/y` expresses combined variation. There are 2 steps involved when solving these equations.

1. Find `k`, by using all other information given. If there are 2 other variables, then values for those variables must be provided.
2. Using the value of `k` that you calculated, and just one of the other variables with NEW values, you can solve for the unknown variable.

Examples 1, 2 and 3 in the textbook walk you through the steps as well.

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Algebra 1a: Chapter 9, Lesson 5, page 417 (day #2).

Inequalities in 2 variables

Given an inequality, treat it as an equality and using the `x` and `y` intercepts, find the solution to the equality. Plot it on your graph paper.

• If the inequality is just a `<`, or `>` problem, then the boundary line (the line you draw connecting the dots) will itself be dotted or dashed. This mean that the points on the line are NOT part of the solution.
• If the inequality has a `≤` or `≥`, then the line will be solid, signifying that the line is part of the solution.

There are 2 ½ planes on the graph, one side of the boundary line that belongs to the solution set (this side will be shaded as part of the solution) and the other side of the line that does not satisfy the inequality.

Now to figure out what ½ plane to shade, pick a point [I like to pick `(0, 0)` or `(1, 1)`] and try those `(x, y)` values in the inequality.

• If the point chosen makes the inequality TRUE, then shade that part of the plane.
• If the point chosen does not satisfy the inequality, then shade the OPPOSITE side ½ plane.

The textbook is actually pretty good in this area, see pages 417-419 for good examples. Purplemath.com has these examples as well.

Two of tonight’s homework problems solved by MrE are here! Just click it

## Day 163 – May 11

Algebra: Chapter 12, Lesson 6, page 561.

Inverse Variation

An equation of the form `y = k/x` where `k` is a constant, expresses inverse variation. Just like in direct variation, find the constant of variation, `k`, then use that to see what other variations can be derived from that constant.

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Algebra 1a: Chapter 9, Lesson 5, page 417.

Inequalities in 2 variables

Given an inequality, treat it as an equality and using the `x` and `y` intercepts, find the solution to the equality. Plot it on your graph paper.

• If the inequality is just a `<`, or `>` problem, then the boundary line (the line you draw connecting the dots) will itself be dotted or dashed. This mean that the points on the line are NOT part of the solution.
• If the inequality has a `≤` or `≥`, then the line will be solid, signifying that the line is part of the solution.

There are 2 ½ planes on the graph, one side of the boundary line that belongs to the solution set (this side will be shaded as part of the solution) and the other side of the line that does not satisfy the inequality.

Now to figure out what ½ plane to shade, pick a point [I like to pick `(0, 0)` or `(1, 1)`] and try those `(x, y)` values in the inequality.

• If the point chosen makes the inequality TRUE, then shade that part of the plane.
• If the point chosen does not satisfy the inequality, then shade the OPPOSITE side ½ plane.

The textbook is actually pretty good in this area, see pages 417-419 for good examples. Purplemath.com has these examples as well.

Two of tonight’s homework problems solved by MrE are here! Just click it

## Day 162 – May 10

Algebra: Chapter 12, Lesson 5, page 557.

Direct Variation

An equation of the form `y = kx` where `k` is a constant, expresses direct variation. `k` is called the constant of variation. Sometimes you have to solve for the constant of variation given the initial information and then use the `k` to solve additional problems given the same `k`.

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Algebra 1a: Chapter 9, Lesson 4, page 413 (day #2).

Inequalities and Absolute Value

If the inequality with absolute values looks like: `| A | < b`, then we solve the conjunction `-b < A < b`. Think of a number line, and the solution will be within the bounds of `-b` and `b`. This also works with `≤`.

If the inequality with absolute values look like: `| A | > b`, then we solve the disjunction `A < -b` OR `A > b`. On the number line, these solutions look like arrows on the outside of the values `-b` and `b`. This works for `≥` as well.

Two of tonight’s homework problems solved by MrE are here! Just click it

## Day 161 – May 9

Algebra: Chapter 12, Lesson 3, page 547.

Linear Functions

A function `f` defined by an equation of the form `y=mx+b`, where `m` and `b` are real numbers, is a LINEAR function and can be written `f(x)=mx+b`.

If that is too confusing, forget the `f(x)` and just substitute `y` instead. Solve for the `f(x)` given the values specified in the domain (the `x` values) .

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Algebra 1a: Chapter 9, Lesson 4, page 413.

Inequalities and Absolute Value

If the inequality with absolute values looks like: `| A | < b`, then we solve the conjunction `-b < A < b`. Think of a number line, and the solution will be within the bounds of `-b` and `b`. This also works with `≤`.

If the inequality with absolute values look like: `| A | > b`, then we solve the disjunction `A < -b` OR `A > b`. On the number line, these solutions look like arrows on the outside of the values `-b` and `b`. This works for `≥` as well.

Two of tonight’s homework problems solved by MrE are here! Just click it

## Day 159 and 160 – May 7,8

Algebra 1: CST Days #6-7  – classroom work days!

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Algebra 1a: CST Days #6-7  – classroom work days!

## Day 154 to 158 – April 30, May 1 – 4

Algebra 1: CST Days #1-5  – classroom work days!

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Algebra 1a: CST Days #1-5  – classroom work days!

## Day 153 – April 27

Algebra 1: CST Review – Work Day!

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Algebra 1a: CST Review – Work Day!

## Day 152 – April 26

Algebra 1: CST Review – Standards 19, 20, 21 and 23 Review Packet (day #2)

36 Problems, use the textbook and your notes and your BRAIN. You can do these!

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Algebra 1a: CST Review Packet – Page 5 and 6

Problems 41 to 54 use the textbook and your notes and your BRAIN. You can do these!