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 2 years 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
 2 years 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

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swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Its a simple question but the wording is confusing me. Maybe u can help me out

JakeV8
 2 years ago
Best 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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Ya it was someone else's question and its bothering me

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0The wording is very wierd

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0It's like the first line I wrote... but no kidding, the wording is awful, probably on purpose...

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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Maybe I am wrong though. I just guessed that. It isnt my question so i wldnt know

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Are 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

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0So, did you happen to see that function earlier that was like h(x,y) = y + c for x<=3 and 5x for x > 3 ?

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Any 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) ?

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0hahah yaaaaa I cldnt figure it out lol

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0or is it a function that might depend on x or y, depending on the x value?

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0ummm its a 3 dimensional function

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0meaning that z depends on x and y

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0The thought I kept having was "What the heck is this" (except shorter, and less printable.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0You think? I mean, it could be, but the question was "is there a value c that makes the function continuous?"

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

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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0idk I have never solved that kind of question b4 but I should look into it

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0often for earlier classes, continuous functions are 2D questions where the point is to evaluate a function for different categories of x values.

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0ohhh I have my calc book with me let me see

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0I assumed it wasn't calc 2... just an early discussion of functions... guess I should have asked what the class was.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0(but now we're curious, aren't we... ;) )

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0lol I forget everything i ever learned. idk after exams i forget everything

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Does that work as a math major?

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0I am majoring in math but for my undergrad

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0My problem is forgetting DURING exams

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

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0If that was true, I was not familiar (nor in agreement!!) with the notation of h(x,y)

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0I mean as a 2D function....

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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Cld be it is a parametric equation that is what it appears to be maybe lol

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0y = 2 is a line, x = 2 is a different line, but I don't like h(x,y) = some hybrid of those 2

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Did you find it? Sorry, I say "2d" out of laziness.. R2

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Guess it didn't save me time typing, but maybe a little thinking :)

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Determine if there is a value for c which makes the function continuous on R2.

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Whoops its def not parametric

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

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Similar to this place, but more math?

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Its more advanced math. I cant answer a single question on there lol

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0oh, sounds scary.... good luck :)

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0lol Not so crazy. I just ask all my questions there

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Determine 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

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

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, you are crazy... but it is fun to help, even if it isn't the best use of limited time... (beats Angry Birds though)

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0hahahah definitely lol

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0I 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!!"

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0lol hahah I dont help that often

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0U 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.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0I 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 :)

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0hahaha. 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

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Sounds 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

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Force for good is NOT equal "spoon feeding 13year old laziness"

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0lol 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

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0So, 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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Is this suppossed to be an eye or a die?

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0or maybe non of the above lol

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0(sigh) now I have to go find a better picture, again. ;)

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Finding an icon pic for OS is a metatimewaster

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0So leave it like this.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0heh... just kidding anyway. I will leave it for now... last search was about 17.437 seconds of effort.

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0I rarely change mine I am way toooo lazy

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Gotta run... catch you in the future... if you get some answer on that h(x,y) thing, let me know.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0but yours is cool and mine looks like an eyeball :)

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0cya gonna start working toooo
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