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The number of State Parks in California is 36 more than one third the number of State Parks in Florida. There are 85 State Parks in California. Which of the following equations would you use to find the number of State Parks in Florida? Note: f represents the number of state parks in Florida.
Quantity f plus 36 all over 3 equals 85
3(f  36) = 85
1 over 3f + 36 = 85
Quantity f minus 36 all over 3 equals 85
 one year ago
 one year ago
The number of State Parks in California is 36 more than one third the number of State Parks in Florida. There are 85 State Parks in California. Which of the following equations would you use to find the number of State Parks in Florida? Note: f represents the number of state parks in Florida. Quantity f plus 36 all over 3 equals 85 3(f  36) = 85 1 over 3f + 36 = 85 Quantity f minus 36 all over 3 equals 85
 one year ago
 one year ago

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swissgirlBest ResponseYou've already chosen the best response.0
Its a simple question but the wording is confusing me. Maybe u can help me out
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
"36 more than" <<< + 36 1/3 of Florida <<< (1/3)F 85 = (1/3)F + 36 Solve for F: 3 ( 8536) = F
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Ya it was someone else's question and its bothering me
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
The wording is very wierd
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
It's like the first line I wrote... but no kidding, the wording is awful, probably on purpose...
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
but option c is \( \large \frac{1} {3f} +36=85 \)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Maybe I am wrong though. I just guessed that. It isnt my question so i wldnt know
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Are you sure that's how it was given, or maybe it was posted erroneously like 1/3f meaning (1/3)f? I've been fighting fraction problems all day... people type without thinking of the possible interpretations, and since it makes all the difference in the world, you have to get clarification
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
So, did you happen to see that function earlier that was like h(x,y) = y + c for x<=3 and 5x for x > 3 ?
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Any idea how to read "h(x,y)" ? Is that a function that depends on both x and y, producing a function output h(x,y) ?
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
hahah yaaaaa I cldnt figure it out lol
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
or is it a function that might depend on x or y, depending on the x value?
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
ummm its a 3 dimensional function
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
meaning that z depends on x and y
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
The thought I kept having was "What the heck is this" (except shorter, and less printable.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
You think? I mean, it could be, but the question was "is there a value c that makes the function continuous?"
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
hmmmm I came across alot of these kind of questions while i was in calc 2 but It wasnt this wording exactly
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Like we never had to make it continous. I am not sure. I am gonna look at the question again
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
idk I have never solved that kind of question b4 but I should look into it
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
often for earlier classes, continuous functions are 2D questions where the point is to evaluate a function for different categories of x values.
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
ohhh I have my calc book with me let me see
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
I assumed it wasn't calc 2... just an early discussion of functions... guess I should have asked what the class was.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
(but now we're curious, aren't we... ;) )
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
lol I forget everything i ever learned. idk after exams i forget everything
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Does that work as a math major?
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I am majoring in math but for my undergrad
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
My problem is forgetting DURING exams
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
like if a function depends on 2 variables then it must be 3d in my opinion but idk
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
I get what you're saying... that could be right... I didn't get the sense we were talking calclevel stuff, so I assumed (maybe wrongly) that it was a weird way of writing a function that was continuous but not "normal" like a simple line.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
If that was true, I was not familiar (nor in agreement!!) with the notation of h(x,y)
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
I mean as a 2D function....
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
but then again the question states in R2 so that means it is 2d
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Cld be it is a parametric equation that is what it appears to be maybe lol
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
y = 2 is a line, x = 2 is a different line, but I don't like h(x,y) = some hybrid of those 2
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Did you find it? Sorry, I say "2d" out of laziness.. R2
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Guess it didn't save me time typing, but maybe a little thinking :)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Determine if there is a value for c which makes the function continuous on R2.
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Whoops its def not parametric
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
ok I can ask the question on MSE and let me see what i get as an answer
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Similar to this place, but more math?
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Its more advanced math. I cant answer a single question on there lol
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
oh, sounds scary.... good luck :)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
lol Not so crazy. I just ask all my questions there
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Determine if there is a value for c which makes the function continuous on R2. h(x,y) = {c + y, if x less/equal 3 { 5x, x>3 Something is wrong with the question though. Like with the brackets
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I am crazzzyyyy I just spent my whole evening here on OS helping when I should have been studying.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Yes, you are crazy... but it is fun to help, even if it isn't the best use of limited time... (beats Angry Birds though)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
hahahah definitely lol
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
I have become much less productive, traditionally measured, since I found this site. But I have become considerably happier, and hopefully helping people will have something of a multiplier effect in the bigger picture. That isn't much of a story to tell a prof the next day when you bomb an assignment "cuz I was helping!! HELPING, I say!!"
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
lol hahah I dont help that often
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
U get sick of this site at some point. I am nearly on here a year. Of late I have been coming on but like you get bored of helping. Sattellite has been her for forever and never gets sick of it so maybe you wont.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
I think pacing yourself may be key... just because it's online and fun for a minute doesn't mean doing the same algebra 1 problem repeatedly for a year is also fun. I can already tell I'm getting more selective with questions I'm picking... I may not make it a year without rethinking my priorities :)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
hahaha. Well I used to use this site to ask my questions back in the day but now its too advanced so I go elsewhere. I used to have fun with the users back then. Now its a bunch of 13 yr olds back then the users were alot older i think. We all used to have groups
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Sounds more fun than now... I don't mind helping a 13year old home school or alt. school student who honestly cares, but it seems like about 50% are just throwing up their hw (or worse, exam questions) and expecting someone to hand them an answer
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Force for good is NOT equal "spoon feeding 13year old laziness"
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
lol I agree. I sometimes just give ppl answers and then afterwards I am like y did i just do that. Today I gave away like 7 answers to this one guy. He kept on asking i and i kept answering
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
So, just prior to signing off, thought it was time for a change of profile pic... kept getting grief about V8 juice, and the v8 doesn't mean that anyway (doesn't really mean anything). We'll see how this new pic works out... gives me an easy answer if I don't know how to solve a question... I'll just say "Ask again" or "It depends" in a dreamy bluish liquidy voice
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
Is this suppossed to be an eye or a die?
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
or maybe non of the above lol
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
(sigh) now I have to go find a better picture, again. ;)
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Finding an icon pic for OS is a metatimewaster
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
So leave it like this.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
heh... just kidding anyway. I will leave it for now... last search was about 17.437 seconds of effort.
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I rarely change mine I am way toooo lazy
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
Gotta run... catch you in the future... if you get some answer on that h(x,y) thing, let me know.
 one year ago

JakeV8Best ResponseYou've already chosen the best response.0
but yours is cool and mine looks like an eyeball :)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
cya gonna start working toooo
 one year ago
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