anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
seb.cal
  • seb.cal
Can I get a medal?
anonymous
  • anonymous
@mathmale @Owlcoffee @whpalmer4 I don't have the slightest clue on how to do this please help.
whpalmer4
  • whpalmer4
Do you know how to graph those two functions?

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anonymous
  • anonymous
@whpalmer4 no idea
anonymous
  • anonymous
@whpalmer4 actually i do
anonymous
  • anonymous
negavtive slope
anonymous
  • anonymous
|dw:1449883313181:dw|
whpalmer4
  • 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
  • anonymous
the other function? -5,0?
whpalmer4
  • whpalmer4
The g(x)
anonymous
  • anonymous
thats what i did i drew the gx 0,0 1,-1 2,-4 and 3.-9
whpalmer4
  • whpalmer4
okay, then can you draw f(x) on there?
whpalmer4
  • whpalmer4
your graph isn't really to scale the way you have drawn it if that is g(x)...
anonymous
  • anonymous
...im so confused...
whpalmer4
  • whpalmer4
But maybe making the graph here isn't the important part, let me provide one.
1 Attachment
anonymous
  • anonymous
thank you.
whpalmer4
  • 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
  • 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
  • anonymous
oh i think im getting it now
anonymous
  • anonymous
domain is the x and range is the y
whpalmer4
  • 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
  • anonymous
0 1 2 3
anonymous
  • anonymous
0 -1 -4 -9
whpalmer4
  • whpalmer4
Okay, important safety tip: always label your results.
anonymous
  • anonymous
sorry
anonymous
  • anonymous
I will remember that
whpalmer4
  • 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
  • anonymous
domain is the f(x) and the second is g(x)
whpalmer4
  • whpalmer4
we need the range of f(x) before we go on to g(x)
whpalmer4
  • whpalmer4
and the second is what of g(x), domain or range?
anonymous
  • anonymous
g(x) is the range
whpalmer4
  • whpalmer4
good answer format: "the domain of f(x) is " "the domain of g(x) is " etc.
anonymous
  • anonymous
okay.
anonymous
  • anonymous
so..
whpalmer4
  • whpalmer4
so, as the problem says: List the domain and range for both f(x) and g(x).
anonymous
  • anonymous
okay thanks alot.
whpalmer4
  • 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
  • anonymous
0,-1,-4,-9
whpalmer4
  • 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
  • whpalmer4
Remember, I'm asking for the DOMAIN, which is the range of X values...
whpalmer4
  • whpalmer4
sorry, shouldn't use the word range in the request! :-)
anonymous
  • 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
  • whpalmer4
is this not math that we are doing?
whpalmer4
  • whpalmer4
Why don't we finish this first?
anonymous
  • anonymous
its so confusing g(x) and f of x be honest will I ever use this in life?
whpalmer4
  • 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
  • anonymous
16
whpalmer4
  • whpalmer4
okay, so in high school. what do you plan to do in your life?
anonymous
  • anonymous
medical field
whpalmer4
  • 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
  • anonymous
well i guess ill think of a new job then!
whpalmer4
  • 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
  • anonymous
i see what you mean...
anonymous
  • anonymous
Can you help me with two more problems please
whpalmer4
  • whpalmer4
let's find out :-)
anonymous
  • 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
  • whpalmer4
Do you know what a compound inequality looks like?
anonymous
  • anonymous
yea
anonymous
  • anonymous
7+4x=56y-9
anonymous
  • anonymous
example^
whpalmer4
  • 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
  • anonymous
okay
whpalmer4
  • 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
  • anonymous
mhmm
whpalmer4
  • 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
  • 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
  • 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
  • whpalmer4
did you ever answer the previous one?
anonymous
  • anonymous
yes i think it was 82.4 \[82.4 \ge f \le 659.2\]
anonymous
  • anonymous
ignore the first 82.4
whpalmer4
  • 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
  • anonymous
oh
anonymous
  • anonymous
but i got the expression right? just mixed up?
whpalmer4
  • 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
  • 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
  • anonymous
and i am taking notes...
anonymous
  • 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
  • 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
  • anonymous
nope
anonymous
  • anonymous
wait you said not so it is a legal weight
whpalmer4
  • 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
  • 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
  • anonymous
\[16 \ge w \ge 14\] i believe i am wrong
whpalmer4
  • whpalmer4
No, that works!
anonymous
  • anonymous
it does?
whpalmer4
  • whpalmer4
Yes! You could also flip it around to \[14 \le w \le 16\]completely equivalent
anonymous
  • anonymous
Do you mind helping with more I'm changing my mind on math.
whpalmer4
  • whpalmer4
sure, I have a bit more time..
anonymous
  • anonymous
oh its okay if you need to go
whpalmer4
  • whpalmer4
I don't have to yet, but I will in half an hour or less...
anonymous
  • anonymous
okay
anonymous
  • anonymous
thats a link to the picture didnt realize how big it was
whpalmer4
  • whpalmer4
uh, that didn't work :-)
anonymous
  • anonymous
darn
whpalmer4
  • whpalmer4
looks like you tried to paste an image instead of a link
anonymous
  • anonymous
if i paste the link it wil lead to my account
whpalmer4
  • whpalmer4
do you know to do a screen capture? You could attach the file
anonymous
  • 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
  • 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
  • anonymous
f(t) = 210 + 40*t
anonymous
  • anonymous
am i right?
whpalmer4
  • whpalmer4
yes, very good
anonymous
  • anonymous
thanks !
whpalmer4
  • whpalmer4
how about the other one?
anonymous
  • anonymous
last one its a small problem and than you can go
anonymous
  • anonymous
A=1/2bh
anonymous
  • anonymous
oh we didnt finish it did we
anonymous
  • anonymous
@whpalmer4 are you here?
anonymous
  • anonymous
@whpalmer4 ?
anonymous
  • anonymous
@whpalmer4
anonymous
  • anonymous
@rainbow_rocks03 A=1/2bh
rainbow_rocks03
  • rainbow_rocks03
Ok, can you draw me the equation you need to solve?
anonymous
  • anonymous
\[A=\frac{ 1 }{ 2 }bh\]
whpalmer4
  • whpalmer4
sorry, was occupied elsewhere no, you need another equation for the second gym membership
anonymous
  • anonymous
oh thats okay
whpalmer4
  • whpalmer4
and then we need to find the point where those two intersect
anonymous
  • anonymous
so make a graph?
whpalmer4
  • whpalmer4
you could do it that way, but algebra is fast and easy for this, too
whpalmer4
  • whpalmer4
"Another gym has no joining fee and costs $70 per month. " g(t) = ?
anonymous
  • anonymous
70!
whpalmer4
  • 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
  • whpalmer4
Any question about that?
anonymous
  • anonymous
nope just the last part After _________months, Casey would have paid $________at either gym.
whpalmer4
  • 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
  • whpalmer4
I will make a graph in meantime.
anonymous
  • anonymous
sure
whpalmer4
  • whpalmer4
1 Attachment
anonymous
  • anonymous
yay
anonymous
  • anonymous
\[A=\frac{ 1 }{ 2 }bh\]
anonymous
  • anonymous
thats my last one
whpalmer4
  • whpalmer4
so let me see your work for that health problem. what did you find for how many months?
anonymous
  • anonymous
7
whpalmer4
  • whpalmer4
show me your work
anonymous
  • anonymous
7 months*
anonymous
  • anonymous
okay. \[210+40t=70t\]
anonymous
  • anonymous
subtract 40t from 70t you get 30
whpalmer4
  • whpalmer4
no, you get 30t, not 30
anonymous
  • anonymous
divide 210 by 30 u get 7
anonymous
  • anonymous
t =7
anonymous
  • anonymous
my bad
whpalmer4
  • whpalmer4
\[210+40t=70t\]\[210 +40t-40t = 70t-40t\]\[210=30t\]\[\frac{210}{30} = \frac{30t}{30}\]\[7=t\]
anonymous
  • anonymous
that what i got kinda
whpalmer4
  • whpalmer4
okay, what's the last problem? then I have to split...
anonymous
  • anonymous
\[A=\frac{ 1 }{ 2 }bh\]
whpalmer4
  • whpalmer4
yes, that's the equation for the area of a triangle. but what's the problem to be solved?
anonymous
  • anonymous
that is the problem there isn't a triangle or anything just solve the equation below
anonymous
  • anonymous
solve for h
whpalmer4
  • whpalmer4
okay, well, that's a pretty important detail there, "solve for h"
whpalmer4
  • 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
  • anonymous
divide h
whpalmer4
  • 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
  • whpalmer4
you have \(\frac{1}{2}b\) multiplying \(h\). You just want \(h\). how do you undo the multiplication?
anonymous
  • anonymous
division because its opposite
whpalmer4
  • whpalmer4
okay, why don't you get rid of the \(b\) first?
whpalmer4
  • whpalmer4
what will your equation look like if you divide both sides by \(b\)?
anonymous
  • anonymous
then we have 1/2
whpalmer4
  • 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
  • anonymous
\[a/b=\frac{ 1 }{ 2 }/b\]
whpalmer4
  • whpalmer4
what happened to \(h\)?!?
anonymous
  • anonymous
i dont know
whpalmer4
  • 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
  • anonymous
get h by itself
anonymous
  • anonymous
2ab=h
whpalmer4
  • 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
  • anonymous
thanks so much u helped a lot
anonymous
  • anonymous
bye!!
whpalmer4
  • whpalmer4
good night!

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