PRACTICE ALGEBRA CST QUESTIONS

cst prep 1

cst prep 2

cst prep 3

cst prep 4

cst prep 5

cst prep 6

cst prep 7

cst prep 8

cst prep 9

cst prep 10

ANSWERS to CST PREP QUESTIONS

answers alg test

PRACTICE “SKIP INTRO TO ALGEBRA” TEST QUESTIONS

alg readiness practice test

So here are the following pages for your review:

PROBLEMS # 1 -10

Algebra Practice Test pg 1

PROBLEMS # 1 1-19

Algebra Practice Test pg 2

PROBLEMS # 20 -32

Algebra Practice Test pg 3

PROBLEMS # 33-40

Algebra Practice Test pg 4

PROBLEMS # 41-44

Algebra Practice Test pg 5

PROBLEMS # 45-50

Algebra Practice Test pg 6

ax² + bx + c = 0 (where ‘a’ cannot be zero.)

Solving the equation means finding ‘x’ values that make the equation true. These ‘x’ values are called the roots of the quadratic.

Quadratic equations can have 0, 1 or two roots.

The quadratic formula is derived from the general quadratic equation (below) by completing the square.

The general quadratic equation…

ax² + bx + c = 0

has roots…

This formula, known as the ‘quadratic formula’, is actually two formulas. The ‘±’ symbol should be read as ‘plus or minus’, which means that you have to work out the formula twice, once with a plus sign in that position, then again with a minus sign.

The first step is to identify the coefficients ‘a’, ‘b’ and ‘c’ in your quadratic equation, so that you can substitute them into the formula to calculate ‘x’.

For this equation:

x² – 4x – 5 = 0

There is no number written in front of the x² term, but in that case it is helpful to think of the x² term as 1x² , so then:

a = 1, b = -4, and c = -5

Substituting these values into the formula we get:

Simplifying the square root term:

Calculating the square root:

or

There are two x intercepts at x = -1 and 5

The term b² − 4ac is called the discriminant of the quadratic equation, because it discriminates between three qualitatively different cases:

• If the discriminant is zero, this means that the parabola described by the quadratic equation touches the x-axis in a single point.
• If the discriminant is positive, then there are two different solutions for x and this means that the parabola intersects the x-axis in two points. Furthermore, if the discriminant is a perfect square, the roots are — in other cases they may be quadratic irrationals.
• If the discriminant is negative, the parabola does not intersect the x-axis at all.

ALGEBRA HOMEWORK ASSIGNMENT FOR MARCH 13

march 13 alg hw

HOMEWORK ASSIGNMENT ANSWERS

march 13 alg hw answers

MARCH 6 HOMEWORK ASSIGNMENT

march 6 alg hw

HOMEWORK ASSIGNMENT ANSWERS

march 6 alg hw answers

march 6 pre alg reveiw 2

march 6 alg comp sq 1

HOMEWORK ASSIGNMENT ANSWERS FOR MARCH 6

march 6 alg comp sq 2

HOMEWORK REVIEW OF EXPONENTS AND DISTRIBUTIVE PROPERTY

feb 29 pre alg hw

HOMEWORK ANSWERS

feb 29 pre alg hw answers

Problem #4 from your homework

A flare is launched at 95 feet per second (ft/sec) from a height of 6 feet. The equation for the object’s height h at time t seconds after launch is h = –16t² + 95t + 6, where h is in feet. When does the object strike the ground?

What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. So I’m looking for the time when the height is h=0 . I’ll set h equal to zero, and solve:

0 = –16t² + 95t + 6   (factor out a negative 1)
0 = - 1( 16t² – 95t – 6) (factor into 2 binomials)
0 = -1 (16t +1)(t – 6)  (look for the zeros)

Then t = -1/16 or t = 6. The first solution, -1/16 seconds  before launch, which doesn’t make sense in this context. (It makes sense on the graph, because the line crosses the x-axis at -1/16, but negative time won’t work in this word problem.)

The flare strikes the ground six seconds after launch.

SOLVING QUADRATIC WORD PROBLEMS ASSIGNMENT:

feb 29 alg hw

ANSWERS TO THE WORD PROBLEMS

feb 29 alg hw answers

“Distance” word problems, involve something traveling at some fixed and steady “rate” or “speed”, or else moving at some average speed. Whenever you read a problem that involves “how fast”, “how far”, or “for how long”, you should think of the distance equation, d = rt, where d stands for distance, r  stands for the (constant or average) rate of speed, and t  stands for time.

Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to the correct units.

Here is an example of a typical problem involving distance and the formula d = rt.  The problem will have something to do with objects moving at a constant  of speed or an average rate of speed.

Example:  If you have a word problem that is looking for how far did they go, or how far did they travel, etc., then you are looking for the DISTANCE.  All you have to do is multiply the rate (in mph, ft per sec, etc) and the time (hours, sec, etc.) together to get the distance.

Rate * Time = Distance       60 mph for 4 hours = 60 *4  = 240 miles

55 mph for 3 hours and then they went 60 mph for 5 hours.  What is the total distance?

(55*3) + (60*5) = distance

165 miles + 300 miles = 465 total miles

If you have a word problem that already gives you the DISTANCE and maybe either the rate or time, all you have to do to solve the problem is divide the DISTANCE by either the rate or time.

Example:  You travel 600 miles in 8 hours, what is your rate of speed.  You know the distance 600 miles, so divide it by the time, 8 hours and you will get miles per hour which is the rate of speed.

600 miles / 8 hours = 75 mph

Sometimes you are given the Distance and the rate of  speed and you are asked for how long will it take to reach your destination.

Example:  You will travel 500 miles at a rate of speed of 40 mph, how long will it take to get there?

500 miles/ 40 mph = 12.5 hours

SO HERE IS THE SIMPLE RULE:  If you are looking for the DISTANCE then multiply the numbers.

If you already have the DISTANCE, divide the distance  by either the rate or time.

HOMEWORK FOR FEBRUARY 27

feb 27 pre alg hw

HOMEWORK ANSWERS FOR FEBRUARY 27

feb 27 pre alg hw answers

HOMEWORK FOR FEBRUARY 28

feb 28 pre alg hw

HOMEWORK ANSWERS FROM FEBRUARY 28

feb 28 pre alg hw answers

Okay, this one is already factored for you. But how do I solve this?

Think: If I multiply two things together and the result is zero, what can I say about those two things? I can say that at least one of them must also be zero. That is, the only way to multiply and get zero is to multiply by zero. This is sometimes called “The Zero Factor Property” or the Zero’s or find the x intercepts.

If the product equals zero, you must always have the equation in the form “(quadratic) equals (zero)” before you can attempt to solve it.

The Zero Factor Property tells you that at least one of the factors must be equal to zero. Since at least one of the factors must be zero, I’ll set them each equal to zero:

x – 3 = 0   or   x – 4 = 0

This gives me simple linear equations, and they’re easy to solve:

x = 3  or  x = 4

And this is the solution they’re looking for:  x = 3, 4

Solve x(x + 5) = 0

A very common mistake that students make on this type of problem is to “solve” the equation for “x + 5 = 0″ by dividing off the x. But you can’t divide by zero; dividing off the x makes the implicit assumption that x is not zero. There is no justification for making that assumption!

Even though you are used to variable factors having variables and numbers (like the other factor, x + 5), a factor can contain only a variable, so “x” is a perfectly valid factor. So set the factors equal to zero, and solve:

x(x + 5) = 0
x = 0  or  x + 5 = 0
x = 0  or  x = –5

Then the solution to x(x + 5) = 0 is x = 0, –5

HOMEWORK FOR FEBRUARY 27

feb 27 alg hw

HOMEWORK ANSWERS

feb 27 alg hw answers