# Compass College Algebra Math Test Prep Part I Placement test ACT community college

good day students and welcome to the compass

practice test, sample questions college algebra. i’m going to be going over the release questions,

they are available online. so let’s go ahead and take a look at question 1, there are nine

questions, in this, on the release questions, i’m going to go over all of them. so number

one says what is the next term in the geometric sequence 16, -4,1,-1/4? now for geometric

sequences, the first, the next term differs from the other by something called the common

ratio ok, so this is the general form for writing a list of geometric sequences. so

generally it can be expressed as A1 and then the next one is a1 times r as a2 and then

a1 times r squared and then a3 i mean a1 times r to the third, and then it goes on and on

and on like that. so the whole idea behind geometric sequence is that every time you

multiply by a number called the common ratio, so to go from the first term to the second

term you multiply by r and then to get from the second one to the third you multiply by

r, notice you keep multiplying by exactly the same thing, ok? so that is exactly what’s

happening here, where multiplying by the same number. ok? so from here to here i multiply

by a certain number and then from negative 4 to one i need to multiply by a certain number

and then the same process here and when i figure out what that number is, when i multiply

negative one fourth by that number, that will tell me what the answer is, so that number

is called the common ratio alright? so how do you find the common ratio? to find the

common ratio all you simply do is you can divide the second term and the first term

or the third term by the second term or the fourth term by the third term the general

formula is simply an divided by an minus 1 so term divided by the term before it to give

you the common ratio alright? so in this problem to find the common ratio i’m just going to

pick this is a1, a1 is 16 and a2 is -4 so to find the common ratio, i’ll, let’s just

use this a2 and a1, i could use a3 and a2 or a4 and a3, it doesn’t matter, you get exactly

the same ratio, ok? so to get your common ratio i’ll divide a2 which is -4 by a1 which

is 16, if you divide that out you get, reduces to -1/4. so what on earth does this mean,

this means that every single time i’m multiplying by -1/4 ok? so we multiply 16 by -1/4 you

get -4, when you multiply -4 by -1/4 guess what you get 1, and then when you multiply

1 by -1/4 you get -1/4 and to get the next term just simply multiply by -1/4 again, ok?

so to get the answer to this problem, what i’m going to do , see i’m looking for a5,

so to get a5, the fifth term or the next term, i’m just going to take a4 which is -1/4 and

i’m going to multiply by what? the common ratio which is -1/4 ok? so what is -1/4 times

-1/4? just multiply across, top to top 1, minus and minus is a plus, 4 times 4 is 16,

so your final answer for number 1 is option c. ok? so there you have it. now let’s take

a look at number 2. it says a manufacturing company processes raw ore. The number of tons

of refined material the company can produce during t days using Process A is A of t equals

t squared plus 2t and using Process B is b of t equals 10t. the company has only 7 days

to process ore and must choose the processes. What is the maximum output of refined material,

in tons for this time period? so we have two processes here represented by two different

processes. so the question is which of these two functions will produce the biggest output,

the maximum output. So there are two approaches that we can use to solve this problem. we

can use the graphical approach and compare the graphs, think about what the graphs of

these two look like as they approach infinity, or we can use a numerical approach ok? so

these are the functions that we get a of t and b of t, now um i’m going to use a numerical

approach for this problem because it’s less complicated, the graphical approach um is

a little bit more evolved, i’m trying to keep it easy here so that you do well on your compass

test ok? so if you can use a numerical approach anytime, it’s good to just make sure that

you see exactly what’s going on ok? alright so what I’m going to do is I’m going to generate

a table of values, and i’m going to compare the output of the values of these two, um

for the seven days and see which one generates the maximum output ok? alright so lets go

ahead and do it. alright so i’m going to start out with the quadratic function, to make a

table of values and then complete the output value, so for the quadratic function we’re

going to have t column here and then um a of t equals t squared plus 2t. so we’re going

to start out from 1, time is unidirectional just goes positive, it doesn’t go negative,

so we’re going to go from 1 all the way to 7. ok so let’s start from 1, so from 1 it’s

a of 1 which is going to be 1 squared plus 2 times 1, what did i just do? I just plugged

in 1 for t here and here, again i want to see what the output is going to be. so it’s

going to be 1 plus 2 which is 3, ok? and then 2 a of 2 and then i have 2 squared plus 2

times 2 which is 4 plus 4 and the output for 2 days, after 2 days is 8. alright so 3, a

of 3, you’re going to have 3 squared plus 2 times 3, that’s going to be 9 plus 6 and

the output for that is 15. for day 4, a of 4 is 4 squared plus 2 times 4 which is 16

plus 8 and the output is 22. 5, a of 5 is 5 squared plus 2 times 5 which is 25 plus

10 the output is 35, the reason i’m dong this is because you have to be really careful with

quadratic functions because they can change direction, um in this case the vertex is in

the negative area so we’re going to be going in the same direction, but you never know,

if you don’t know what the graph looks like you can’t make assumptions as to what the

pattern, the direction of the change in magnitude is. so let’s finish this up, A of 6 you’re

going to have 6 squared plus 2 times 6 which is 36 plus 12, the output after 6 days is

48 and then for 7, you have a of 7, just input 7 into your function and you have 7 squared

plus 2 times 7 which equals 49 plus 14 and the answer is 63 ok? so you can see that the

output is increasing and the maximum output using this process A is 63 after seven days.

So the second process is a linear process, it’s much easier to generate a table of values,

this one you can just simply plug in 7 since it’s unidirectional, i just want to show you

the pattern ok? so for b of t is 10 t, because if you notice that’s what it says here, b

of t equals 10t so that’s what i’m using alright? so we’re going to start from 1 , this one

you just simply multiply by 10, so b of 1 is 10 times 1 which is 10. And then for 2

b of 2 is 10 times 2 which is 20. And the pattern continues, b of 3 is 10 times 3 which

is 30. And then b of 4 output after 4 days is 40. 5 days, output after 5 days is 50,

6 output after 6 days is 60. and 7 output after 7 days is 70 ok? so I just wanted, you

to see the comparison here, alright? compare them you can see that for these two the outputs

keep increasing, the biggest output for the process A is 63 and the biggest output for

process B is 70 so the winner here is process B at the maximum output after 7 days, up to

7 days is 70 and the answer is E ok? Alright so let’s move on to question number 3, it

says for the 2 functions, f of x and g of x, tables of values are shown below. What

is the value of g of f of 3? ok so g of f of 3 just means that, the value gets inputted

into a function, and then the value of that function gets inputted into another function

alright? so basically starting with x and then um x is inputted into the function f

and when you plug in x into f you end up with f of x. and then f of x gets inputted into

another function, g and then your final output is going to be g of f of x ok? so we’re just

going to do this step wise first thing you take our input plug it into the first function

to get our first output. and what ever output we generate we’ll plug it into our second

function to get our final result which is g of f of 3 ok? alright so let’s go ahead

and do it, so we’re going to start with x, x is um what is x? the innermost value which

is 3, that’s what x is, x is 3 ok? so we got to first of all look for f of 3, so you gotta

use the table that relates, f and x, this is the table that’s going to make this connection

right here, so we look for 3, there goes 3 in the x what is the output, the output for

3 is 2 so f of 3 is 2 ok? so now we know that from 3, this function has taken us to the

value of 2 now we’re going to use this, this 2 now becomes an input ok? and that becomes

our new x so this will be inputted into g to get what the output is, so now we’re going

to look for the chart that relates an input for the function g, so now we’re going to

look for g of f of 3 ok? but wait a minute though, what is f of 3? f of 3 is 2 so what

we’re looking for in essence is g of, let me put it in blue, g of 2 ok? so g of 2 we’re

going to look for an input under x an g table and see what the output is when the input

is 2, so there you have it right here so when x is 2 what is the output? the output is -3

alright so g of 2 is -3. so there goes your final answer, your final answer for number

3 is B ok? alright let’s go ahead and move on to question number 4. Question 4 says for

positive real numbers x,y, and z, which of the following expressions is equivalent to

x to the 1/2 y to the 2/3 and z to the 5/6. now if you look at this um we need to make

all the denominators the same so that we can have a common root. then we extract that root,

so in order to make all the denominators the same we need to take a step back to middle

school and remember the whole idea of lowest common denominator, alright? so the question

is what is the lowest common denominator of 2, the first denominator, 3 the second denominator

and 6. the LCD of all this, of these three numbers is 6 ok? so the goal is to make all

the denominators have the same number, namely the lowest common denominator, make them the

lowest common denominator and then we’re going to extract it and that will be our common

root ok? so let’s do it, so we have x to the 1/2, i need this to be 6, so what do i multiply

it by? i times it by 3, top and bottom, and then for the y component, y to the 2/3 i need

this to be 6 so i multiply it by 2 top and bottom. and then for z 5/6 i need this to

be 6 on the bottom, it’s already 6 so i multiply it by 1 or i could just leave it alone, it

makes no difference alright? so now this becomes, x to the 3/6 y to the 4/6 and z to the 5/6,

ok so this is exactly what we want. We want all of the denominators to be identical and

that’s the case here. so since we have out identical denominator watch this, i’m going

to use the properties of exponents to extract the denominator from all these um three powers

here, so we’re gonna have x to the third, y to the fourth, z to the fifth, factoring

out the denominator equals 1/6 ok? because notice if I distribute 1/6 to all these 3

numbers what am I going to get? i’m going to end up with this, ok so I might as well

just factor it out, so remember the rule that x to the 1 over n equals the nth root of x

ok? this is the root. alright and the numerator is the power, the power here is 1 so it’s

inconsequential. so i’m going to express this using 6 as the root so we’re going to have,

this going to be counted 6 root of x to the 3rd, y to the 4th and z to the 5th ok? so

what answer is that? 3,4,5 the answer is D. alright, now let’s move on to question number

5 this involves operation of matrices, this is a subtraction. so um this question might

look easy but you have to be really careful when you’re carrying out your arithmetic on

the different uh elements in the matrix ok? the different cells in the matrix ok? in this

case you have a subtraction and you also have negative numbers, so you have to be careful

when resolving the signs alright? so i’m going to set it up, A minus B like this, this is

matrix A on the left and matrix B is on the right, you must preserve the order whenever

you carry any operations with matrices because it’s very sensitive to order, matrices are

very sensitive to order of operations ok? Alright so to do that we’re going to be subtracting

B from matrix A so, preserve the order of the elements, we’re going to go 2 minus , don’t

forget that sign there that’s a trap, negative 2 for the first column, first row. first row

second column we’re going to have negative 4 minus 4 ok that completes my first row.

Second row first column we’re going to have 6, minus negative 6 and then for our second

row second column we’re going to have zero minus zero. Now let’s use our arithmetic skills

to finish this off. these two are minuses when you subtract a minus and minus you multiply

the signs so they become plus, so 2 plus 2 is 4, minus 4, minus 4 is negative 8, 6 minus

6, these are two minuses you multiply to be a plus, 6 plus 6 is 12, 0 minus 0 is 0. so

you have 4 negative 8, 12 and 0, answer is option E ok? So thanks so much for paying

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Thanks again and have a wonderful day.