Algebra 1: CST Review – Standards 9, 10, 12, 13, 14 and 15 Review Packet (day #2).
36 Problems, use the textbook and your notes and your BRAIN. You can do these!
======================================================================
Algebra 1a: CST Review Packet – Page 1 and 2.
Problems 1 to 29 use the textbook and your notes and your BRAIN. You can do these!
Algebra 1: CST Review – Standards 9, 10, 12, 13, 14 and 15 Review Packet (day #1)
36 Problems, use the textbook and your notes and your BRAIN. You can do these!
======================================================================
Algebra 1a: CST Review – Volume of Prisms and Cylinders
Its easy to do the volumes of prisms, find the area of the base and then multiply that by the height of the prism. For a rectangular prism, find the base area, `A=l⋅w`, then multiply that by the height `h`, or Volume, `V=l⋅w⋅h`.
For a cylinder, just replace the area of the base with the area of a circle, `A=πr^2`, then multiply that by the height. So, for a cylinder, we can rewrite that as `V=(πr^2)(h)`.
Algebra 1: CST Review – Standards 2, 4, 5, 6 and 7 Review Packet (day #2)
58 Problems, use the textbook and your notes and your BRAIN. You can do these!
======================================================================
Algebra 1a: CST Review – SURFACE AREA
Don’t forget SURFACE AREA of RECTILINEAR PRISMS and CYLINDERS!
For cubes and rectangles: AREA, `A = l * w` and multiply by `2`, the number of IDENTICAL FACES and then add them up.
For cylinders, find the lateral area (the side) and the lids:
AREA of LIDS: Area, `A = π* r^2`, there are usually 2 of these
LATERAL AREA: The length * width (or height) of the ‘unrolled paper tube’. The length is the circumference of the circle and the width is the height of the tube. So …
LATERAL AREA: `A = h * (2π *r)`
Algebra 1: CST Review – Standards 2, 4, 5, 6 and 7 Review Packet
58 Problems, use the textbook and your notes and your BRAIN. You can do these!
======================================================================
Algebra 1a: CST Review – SURFACE AREA
Don’t forget SURFACE AREA of RECTILINEAR PRISMS and CYLINDERS!
For cubes and rectangles: AREA, `A = l * w` and multiply by `2`, the number of IDENTICAL FACES and then add them up.
For cylinders, find the lateral area (the side) and the lids:
AREA of LIDS: Area, `A = π* r^2`, there are usually 2 of these
LATERAL AREA: The length * width (or height) of the ‘unrolled paper tube’. The length is the circumference of the circle and the width is the height of the tube. So …
LATERAL AREA: `A = h * (2π *r)`
Algebra: Chapter 13, Lesson 7, page 602.
Using Quadratic Equations
Word problems need to be thought out before starting. Draw pictures and understand what information you know and what you need to find out. With DRT problems, you may need to manipulate the `D * R = T` to put it in terms of what you know. Equivalent equations are `R = T/D` and `D = T/R`.
The Pythagorean theorem will also come in handy for these types of problems! REMEMBER, `c^2 = a^2 + b^2`
Once you have the equation, it is probably going to be quadratic and you may have to come up with an LCM. Use all factors in the denominators and you should be fine.
Two of tonight’s homework problems solved by MrE are here! Just click it.
======================================================================
Algebra 1a: CST Review – SURFACE AREA
Don’t forget SURFACE AREA of RECTILINEAR PRISMS and CYLINDERS!
For cubes and rectangles: AREA, `A = l * w` and multiply by `2`, the number of IDENTICAL FACES and then add them up.
For cylinders, find the lateral area (the side) and the lids:
AREA of LIDS: Area, `A = π* r^2`, there are usually 2 of these
LATERAL AREA: The length * width (or height) of the ‘unrolled paper tube’. The length is the circumference of the circle and the width is the height of the tube. So …
LATERAL AREA: `A = h * (2π *r)`
Algebra: Chapter 13, Lesson 6, page 598.
Solving Radical Equations
We can solve radical equations by first using the principle of squaring to find the quadratic equation. Remember, squaring a square root results in the stuff under the radical being displayed without any exponents.
For example the `(sqrt (27 – 3x))^2` is just `27 – 3x`!
If we have formulae and you have to solve it for a given variable, treat that variable as the ONLY one, and treat the other variables as if they are constants. Manipulate the formula to isolate just that variable. See the examples on page 599 for formulae examples. JUST CONCENTRATE ON THE DESIRED VARIABLE AND TREAT THE OTHERS AS IF THEY ARE NUMBERS!
Here are some more examples for purplemath. The pages before and after this one are also pretty good review.
Two of tonight’s homework problems solved by MrE are here! Just click it.
======================================================================
Algebra 1a: CST Review Packet (day #2).
QUIZ PRACTICE #1 – today in class and for homework!
Algebra: Chapter 13, Lesson 5, page 595.
Solving Rational Equations
Remember, we solve rational equations by multiplying both sides by the LCM of all the denominators. This can result in a quadratic equation. Sometimes, there can be extraneous solutions, so … you should always check all possible solutions.
Practice makes perfect, use ANY method to factor:
Here are some more examples too.
Two of tonight’s homework problems solved by MrE are here! Just click it.
======================================================================
Algebra 1a: CST Review Packet
QUIZ PRACTICE #1 – today in class and for homework!
Algebra: Chapter 13, Lesson 4, page 589 (day #2).
The Quadratic Formula, finally!
Given that `ax^2 + bx + c = 0`, then the quadratic formula
`x = (−b±sqrt(b^2−4ac))/(2a)`
gives the solutions of the quadratic equation. This requires that the quadratic equation is always in standard form:
Memorize it and memorize the discriminant, the expression under the radical (the `b^2−4ac` thingy).
Don’t forget that the
all mean the same thing. By definition, the equation `ax^2+ bx + c=0` implies that we are setting `y = 0 ` and finding the x-intercepts or the roots or the answers or the solutions!!
Once again, Purplemath.com comes to the rescue, check out these examples for the use of the quadratic formula!
Two of tonight’s homework problems solved by MrE are here! Just click it.
======================================================================
Algebra 1a: CST Review – Equalities and Inequalities – GRAPHING
In graphing equalities and inequalities in x (number line) and x-y graphs, remember to use a T-chart for x-y graphs and to reverse the sign when graphing inequalities when dividing or multiplying by a negative number.
Algebra: Chapter 13, Lesson 4, page 589.
The Quadratic Formula, finally!
Given that `ax^2 + bx + c = 0`, then the quadratic formula
`x = (−b±sqrt(b^2−4ac))/(2a)`
gives the solutions of the quadratic equation. This requires that the quadratic equation is always in standard form:
Memorize it and memorize the discriminant, the expression under the radical (the `b^2−4ac` thingy).
Don’t forget that the
all mean the same thing. By definition, the equation `ax^2+ bx + c=0` implies that we are setting `y = 0 ` and finding the x-intercepts or the roots or the answers or the solutions!!
Once again, Purplemath.com comes to the rescue, check out these examples for the use of the quadratic formula!
Two of tonight’s homework problems solved by MrE are here! Just click it.
======================================================================
Algebra 1a: CST Review – Area of a Parallelogram, Triangle and Trapezoid
Definitions: `b` is the base and `h` is the height of the geometric figure below. Remember too, that the base and the height are at RIGHT ANGLES to each other!
Area, of a Triangle, `A = (b*h)/2` which is the same thing as `(1/2) * b * h`
Area of a Parallelogram, `A= b*h`.
Area of a Trapezoid, `A = (1/2) * h * (b_1 + b_2)` which is the same thing as `[h * (b_1 + b_2)]/2`
Algebra 1a: CST Review – Equalities and Inequalities – GRAPHING
In graphing equalities and inequalities in x (number line) and x-y graphs, remember to use a T-chart for x-y graphs and to reverse the sign when graphing inequalities when dividing or multiplying by a negative number.
Algebra: Chapter 13, Lesson 3, page 586.
Solving quadratics by completing the squares
We can use the technique of completing the square to solve quadratic equations. Recall that the addition property allows us to add a number to both sides of the equation. To complete a square, take `1/2` of the x-coefficient, square it and add it to both sides of the quadratic equation.
For example:
`x^2 – 4x – 7 = 0` becomes `x^2 – 4x = 7` by adding 7 to both sides.
Now, take `1/2` of the x coefficient (the `-4` in `-4x`). `1/2` of the `-4` is `- 2`. Now square that, `-2^2=4` and
`x^2 – 4x + 4 = 7 + 4` by adding 4 to both sides to complete the square.
REMEMBER: `x^2 – 4x + 4` factored becomes `(x – 2)^2`
`(x – 2)^2 = 11` or by square rooting each side to `x – 2 = ±sqrt(11)` so that
`x = 2 ± sqrt(11)` or `2 + sqrt(11)` and `2 – sqrt(11)` as the final solutions.
Purplemath explains it too, just click here for step-by-step instructions.
Two of tonight’s homework problems solved by MrE are here! Just click it.
======================================================================
Algebra 1a: CST Review – Area of a Parallelogram, Triangle and Trapezoid
Definitions: `b` is the base and `h` is the height of the geometric figure below. Remember too, that the base and the height are at RIGHT ANGLES to each other!
Area, of a Triangle, `A = (b*h)/2` which is the same thing as `(1/2) * b * h`
Area of a Parallelogram, `A= b*h`.
Area of a Trapezoid, `A = (1/2) * h * (b_1 + b_2)` which is the same thing as `[h * (b_1 + b_2)]/2`
