MEDAL/FAN
Let f(x) be a line with slope -5 and y intercept 0 with domain {0, 1, 2, 3}, and let g(x) = {(0, 0), (1, -1), (2, -4), (3, -9)}. Compare the two functions. List the domain and range for both f(x) and g(x). Are the x intercepts the same or different?

- anonymous

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- seb.cal

Can I get a medal?

- anonymous

@mathmale @Owlcoffee @whpalmer4
I don't have the slightest clue on how to do this please help.

- whpalmer4

Do you know how to graph those two functions?

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## More answers

- anonymous

@whpalmer4 no idea

- anonymous

@whpalmer4 actually i do

- anonymous

negavtive slope

- anonymous

|dw:1449883313181:dw|

- whpalmer4

Good. Can you draw the other function on there? (click on the little pencil at the upper right corner and you can draw on top of it)

- anonymous

the other function? -5,0?

- whpalmer4

The g(x)

- anonymous

thats what i did i drew the gx 0,0 1,-1 2,-4 and 3.-9

- whpalmer4

okay, then can you draw f(x) on there?

- whpalmer4

your graph isn't really to scale the way you have drawn it if that is g(x)...

- anonymous

...im so confused...

- whpalmer4

But maybe making the graph here isn't the important part, let me provide one.

##### 1 Attachment

- anonymous

thank you.

- whpalmer4

The problem says there is a function f(x) which goes through (0,0) and has a slope of -5. That's the line in my graph.
The problem also says there is a function g(x) which has the points listed in the table. Those are the blue dots.

- whpalmer4

So now we have some idea of what this stuff looks like, which is often a big help.
" Compare the two functions. List the domain and range for both f(x) and g(x). Are the x intercepts the same or different? "
Do you know the definition of domain and range?

- anonymous

oh i think im getting it now

- anonymous

domain is the x and range is the y

- whpalmer4

domain is the set of possible values of x
range is the set of possible values of y
So, can you list the domain and range for f(x)? The problem actually states the domain for f(x), but not the range.
How about g(x)?

- anonymous

0
1
2
3

- anonymous

0
-1
-4
-9

- whpalmer4

Okay, important safety tip: always label your results.

- anonymous

sorry

- anonymous

I will remember that

- whpalmer4

so the first one is obviously the domain. is it the domain of f(x) or g(x)? the second one is what?

- anonymous

domain is the f(x) and the second is g(x)

- whpalmer4

we need the range of f(x) before we go on to g(x)

- whpalmer4

and the second is what of g(x), domain or range?

- anonymous

g(x) is the range

- whpalmer4

good answer format:
"the domain of f(x) is "
"the domain of g(x) is "
etc.

- anonymous

okay.

- anonymous

so..

- whpalmer4

so, as the problem says:
List the domain and range for both f(x) and g(x).

- anonymous

okay thanks alot.

- whpalmer4

Let me summarize the answer so far:
The domain of f(x) is {0,1,2,3}
Now we need the domain of g(x). Can you tell me what is the domain of g(x)?

- anonymous

0,-1,-4,-9

- whpalmer4

Sigh. What did I say about labeling your answers? I just gave you an example with my statement:
"The domain of f(x) is {0,1,2,3}"
What is the domain of g(x)?

- whpalmer4

Remember, I'm asking for the DOMAIN, which is the range of X values...

- whpalmer4

sorry, shouldn't use the word range in the request! :-)

- anonymous

I make fake accounts on here all the time I'll let you know when it's me just by sending you a message but as of now I need to work on my math.

- whpalmer4

is this not math that we are doing?

- whpalmer4

Why don't we finish this first?

- anonymous

its so confusing g(x) and f of x be honest will I ever use this in life?

- whpalmer4

"let g(x) = {(0, 0), (1, -1), (2, -4), (3, -9)}"
So the domain is the set of allowable x values, right? and the range is the set of y values?
Here we have defined the function by listing all of the values. By convention, when you see (some number, some other number) that means the first number is x and the second number is y
So, the values that g(x) can take on are
0 = g(0)
-1 = g(1)
-4 = g(2)
-9 = g(3)
the domain of g(x) is the x values: {0,1,2,3}
the range of g(x) is the y values: {0,-1,-4,-9}
As for whether you will use this in your life: how old are you?

- anonymous

16

- whpalmer4

okay, so in high school. what do you plan to do in your life?

- anonymous

medical field

- whpalmer4

okay, unless your definition of medical field means janitorial work at the clinic or being the scheduler for appointments or the like, you will have many more math classes in your future, and yes, you will need to know this to pass them, regardless of whether or not you use it while delivering medical care.

- anonymous

well i guess ill think of a new job then!

- whpalmer4

if medicine is what you want to do, then it is inescapable that you'll do a bunch of math, chemistry, etc. while you study to get most professional training for the field. But rather than just chucking aside your plans, maybe buckling down and learning something that may not seem easy is the right thing to do. Just about anything is going to have parts that will be difficult to learn at first.

- anonymous

i see what you mean...

- anonymous

Can you help me with two more problems please

- whpalmer4

let's find out :-)

- anonymous

Enter a compound inequality to show the levels that are within each range.
A typical acoustic guitar has a range of three octaves. When the guitar is tuned to “concert pitch,” the range of frequencies for those three octaves is between 82.4 Hertz and 659.2 Hertz inclusive.
What is the compound inequality that represents the range of frequencies for a guitar tuned to “concert pitch” using the variable f?

- whpalmer4

Do you know what a compound inequality looks like?

- anonymous

yea

- anonymous

7+4x=56y-9

- anonymous

example^

- whpalmer4

Okay, so write a compound inequality that says that the frequency, \(f\), is between 82.4 Hertz and 659.2 Hertz
Oh, no, that's not a compound inequality. An inequality has an inequality sign in it, one of \(< > \ge \le\)
A simple inequality might be
\[\text{age} > 18\]which says that the age is older than 18
A compound inequality has multiple parts. A compound inequality for the years that one is called a teenager might be:
\[12 < \text{age} < 20\]

- anonymous

okay

- whpalmer4

Something important to note in this problem is that word "inclusive" in the description of the frequencies. When you see "inclusive", that means you include the endpoints.
"pick a number between 1 and 5" to a mathematician means choose any number greater than 1 and less than 5. Actually choosing 1 or 5 is not allowed.
"pick a number between 1 and 5, inclusive" to that same mathematician means that now actually choosing 1 or 5 IS allowed.

- anonymous

mhmm

- whpalmer4

"a number \(x\) between 1 and 5"
\[1 < x <5\]
"a number \(x\) between 1 and 5, inclusive"
\[1 \le x \le 5\]

- whpalmer4

So, if you have to write a compound inequality stating that the variable \(f\) is between 82.4 Hertz and 659.2 Hertz, inclusive, what would you write?
If you want to do the combined \(\le\) or \(\ge\) just write <= or >= as appropriate

- anonymous

@whpalmer4 what about this one?
Enter a compound inequality to show the levels that are within each range.
The ball used in a soccer game may not weight more than 16 ounces or less than 14 ounces at the start of the match. At 1 ounce of air was added to a ball, the ball was approved for use in a game.
What was the weight of the ball before adding 1 ounce of air using the variable w?

- whpalmer4

did you ever answer the previous one?

- anonymous

yes i think it was 82.4 \[82.4 \ge f \le 659.2\]

- anonymous

ignore the first 82.4

- whpalmer4

close, but not quite...
you are saying that 82.4 is greater than or equal to f, not that f is greater than or equal to 82.4...
you need
\[82.4 \le f \le 659.2\]

- anonymous

oh

- anonymous

but i got the expression right? just mixed up?

- whpalmer4

well, that is a pretty important part, getting the inequality signs pointing in the right direction. but other than that, if you look at mine, it's the same after fixing that.

- whpalmer4

just remember that when see \(<\) you read it as "is less than"
\(\le\) reads as "is less than or equal to"
\(>\) reads as "is greater than"
\(\ge\) reads as "is greater than or equal to"
and read it as you would say it. You want an expression that says 5 is less than x is less than 10, that's:
\[5 < x < 10\]
Want to say that 82.4 Hertz is less than or equal to the frequency \(f\) is less than or equal to 659.2 Hertz:
\[82.4 \text{ Hertz} \le f \le 659.2\text{ Hertz}\]
if you can say it, you can write it...

- anonymous

and i am taking notes...

- anonymous

how about the other question ut has more than 3 numbers
Enter a compound inequality to show the levels that are within each range.
The ball used in a soccer game may not weight more than 16 ounces or less than 14 ounces at the start of the match. At 1 ounce of air was added to a ball, the ball was approved for use in a game.
What was the weight of the ball before adding 1 ounce of air using the variable w?

- whpalmer4

Well, the weight of the ball is not more than 16 ounces, so does that include 16 ounces as a legal weight, in your opinion?

- anonymous

nope

- anonymous

wait you said not so it is a legal weight

- whpalmer4

well, I think it does: 16 ounces exactly is not more than 16 ounces. 16.1 ounces is more than 16 ounces, but 16 ounces is not. That tells me that we'll have an equals sign as part of our inequality sign (and same for 14 ounces).

- whpalmer4

We know the ball (at the start of the match) is 14 or more ounces and also 16 or fewer ounces, right? Would you agree that is equivalent to what it said?

- anonymous

\[16 \ge w \ge 14\]
i believe i am wrong

- whpalmer4

No, that works!

- anonymous

it does?

- whpalmer4

Yes! You could also flip it around to
\[14 \le w \le 16\]completely equivalent

- anonymous

Do you mind helping with more I'm changing my mind on math.

- whpalmer4

sure, I have a bit more time..

- anonymous

oh its okay if you need to go

- whpalmer4

I don't have to yet, but I will in half an hour or less...

- anonymous

okay

- anonymous

thats a link to the picture didnt realize how big it was

- whpalmer4

uh, that didn't work :-)

- anonymous

darn

- whpalmer4

looks like you tried to paste an image instead of a link

- anonymous

if i paste the link it wil lead to my account

- whpalmer4

do you know to do a screen capture? You could attach the file

- anonymous

Set up and solve a system of equations to solve the problem.
Casey wants to buy a gym membership. One gym has a $210 joining fee and costs $40 per month. Another gym has no joining fee and costs $70 per month. When would Casey pay the same amount to be a member of either gym? How much would he pay?
f(t) =____
+ ________
t
g(t) =_____
t
After
________months, Casey would have paid $
_______at either gym.

- whpalmer4

okay, if \[f(t) = something + something * t\]which of the fees goes where? $210 joining fee and $40/month
can you write out f(t) for me?

- anonymous

f(t) = 210 + 40*t

- anonymous

am i right?

- whpalmer4

yes, very good

- anonymous

thanks !

- whpalmer4

how about the other one?

- anonymous

last one its a small problem and than you can go

- anonymous

A=1/2bh

- anonymous

oh we didnt finish it did we

- anonymous

@whpalmer4 are you here?

- anonymous

@whpalmer4 ?

- anonymous

@whpalmer4

- anonymous

@rainbow_rocks03
A=1/2bh

- rainbow_rocks03

Ok, can you draw me the equation you need to solve?

- anonymous

\[A=\frac{ 1 }{ 2 }bh\]

- whpalmer4

sorry, was occupied elsewhere
no, you need another equation for the second gym membership

- anonymous

oh thats okay

- whpalmer4

and then we need to find the point where those two intersect

- anonymous

so make a graph?

- whpalmer4

you could do it that way, but algebra is fast and easy for this, too

- whpalmer4

"Another gym has no joining fee and costs $70 per month. "
g(t) = ?

- anonymous

70!

- whpalmer4

no. close. You said "70" (I'll ignore the "!" as it has a specific mathematical meaning which you do not intend)
\[g(t) = 70\]
but the function tells us how much the membership has cost up through that point, not how much it costs this month
first one was \[f(t) = 210+40t\]
Join for $210, pay $40 more each month
this gym you have no join fee, but pay $70 per month
\[g(t) = 0 + 70t = 70t\]

- whpalmer4

Any question about that?

- anonymous

nope just the last part
After _________months, Casey would have paid $________at either gym.

- whpalmer4

Now we need to find the point in time at which both memberships will have cost the same amount. Obviously, the f(t) membership is more expensive at first (you have to fork over the $210 join fee) but it costs less per month after that, so the other one will eventually catch up and pass it in total cost.
So we can find that by setting \[f(t) = g(t)\] and solving for \(t\) that makes that true. Can you take a crack at that, please?

- whpalmer4

I will make a graph in meantime.

- anonymous

sure

- whpalmer4

##### 1 Attachment

- anonymous

yay

- anonymous

\[A=\frac{ 1 }{ 2 }bh\]

- anonymous

thats my last one

- whpalmer4

so let me see your work for that health problem. what did you find for how many months?

- anonymous

7

- whpalmer4

show me your work

- anonymous

7 months*

- anonymous

okay. \[210+40t=70t\]

- anonymous

subtract 40t from 70t you get 30

- whpalmer4

no, you get 30t, not 30

- anonymous

divide 210 by 30 u get 7

- anonymous

t =7

- anonymous

my bad

- whpalmer4

\[210+40t=70t\]\[210 +40t-40t = 70t-40t\]\[210=30t\]\[\frac{210}{30} = \frac{30t}{30}\]\[7=t\]

- anonymous

that what i got kinda

- whpalmer4

okay, what's the last problem? then I have to split...

- anonymous

\[A=\frac{ 1 }{ 2 }bh\]

- whpalmer4

yes, that's the equation for the area of a triangle. but what's the problem to be solved?

- anonymous

that is the problem there isn't a triangle or anything just solve the equation below

- anonymous

solve for h

- whpalmer4

okay, well, that's a pretty important detail there, "solve for h"

- whpalmer4

\[a = \frac{1}{2}bh\]to solve for \(h\), you need \(h\) all by itself on one side of \(=\)
how are you going to do that? You can do just about anything you want as long as you do it to both sides of the equation. Add, multiply, divide, subtract, square root, etc.

- anonymous

divide h

- whpalmer4

generally, you add or subtract to get all the terms that do not include your target to the other side, then multiply or divide to get rid of anything that is "attached" to your target

- whpalmer4

you have \(\frac{1}{2}b\) multiplying \(h\). You just want \(h\). how do you undo the multiplication?

- anonymous

division because its opposite

- whpalmer4

okay, why don't you get rid of the \(b\) first?

- whpalmer4

what will your equation look like if you divide both sides by \(b\)?

- anonymous

then we have 1/2

- whpalmer4

no, what does the whole equation look like?
really, you'll get more benefit if you write out the whole thing for each step.

- anonymous

\[a/b=\frac{ 1 }{ 2 }/b\]

- whpalmer4

what happened to \(h\)?!?

- anonymous

i dont know

- whpalmer4

\[A = \frac{1}{2}bh\]let's multiply both sides by 2 to get rid of that fraction.
\[2*A = 2*\frac{1}{2}bh\]\[2A = \cancel{2}\frac{1}{\cancel{2}}bh\]\[2A = bh\] what is my next step?

- anonymous

get h by itself

- anonymous

2ab=h

- whpalmer4

you can't move the \(b\) like that!
divide both sides by \(b\):
\[\frac{2A}{b} = \frac{bh}{b}\]
\[\frac{2A}{b} = \frac{\cancel{b}h}{\cancel{b}}\]\[\frac{2A}{b} = h\]\[h = \frac{2A}{b}\]

- anonymous

thanks so much u helped a lot

- anonymous

bye!!

- whpalmer4

good night!

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