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Here is my work: Part A: xy=A xy=280 x+1=y-5 x= y-6 Part B: (y-6)(y)=280 y^2-6y=280 y^2-6y-280=0 (y+14)(y-20) The width of the land is 20 meters.
have you thought about using "l" and "w" instead of "x" and "y" 'cos you may have got them the wrong way round.
so length would be: l-5 and width would be: w+1 so does that mean I would have to set l-5=w+1?
i dont think they're asking me to solve for x for part A, just to give them an equation...
I think I understand..
A rectangular area is usually denoted with 'L' for length and 'W' for width , since it is a rectangle L and W are not equal -Of course! but the statement then says that if the rectangular area is to be a square under two conditions , That is: i) L-5 ii) W +1 Right ? . So if we think about the square part , all sides are the same . since that is true we can use another variable , lets call it 'X' and we know the area of a square is x*x = Area
so that would be my answer to part A? x= 280/(x+5)?
Do you understand what I did ??
x*x = area of a square however we were given a rectangle but they told us that if one side is L -5 and the other side W+1 would make that rectangular into a sqaure so I said , why not use the the area of the square equation X = L-5 and X = W +1 <--- This should totally make sense , right ?
yes i think so??
area of a rectangle is L*W = area let L = X+5 so (x+5)*W = area finally W = area/(x+5)
okay i understand that!
is there something else i need to do for part a then?
or is that it
@matenrouyes i am sorry if i have added to any confusion i was merely suggesting that you may have forgotten what x and y stand for when you started out, did x stand for length or width. i'd say the latter. so when you conclude rightly that y = 20 are you not concluding that length = 20 ?! that was the only point i was making and that was why i also suggested using more descriptive variable names
so i paste your entire post into a text editor, i switch w for x and l for y and i get this: Part A: wl=A wl=280 w+1=l-5 w= l-6 Part B: (l-6)(l)=280 l^2-6l=280 l^2-6l-280=0 (l+14)(l-20) see my point?
OHHH I SEE!!
cool apart from that, it's excellent
Thank you!! do you mind staying while i work it out to check it again?
OK: so i think that for part a i had to find the equation of L, so this is what i did Part A: L-5=w+1 L= w+6 and then for part b, i would plug in what i got for A to get the width (w+6)(w)=280 and when i solve it i get -20, and 14, and obviously you cant have negative meters so the width is 14?? @IrishBoy123
oops sorry for the wrong answer . (u_u)
yes adding in a few of the details that an instructor might wish to see and that you already included above: \((w+6)w=280\) \(w^2 + 6w -280 = 0\) \((w+20)(w-14) = 0\) \(w = 14\) or \(w = -20\) so \(w = 14\)
but can you agree that what i put for part A is correct?
ok i'd stick with what you did originally a part from the fact you forgot what x and y were representing so i pasted your work into a proessors and switched for w ad l and i got this: Part A: wl=A wl=280 w+1=l-5 w= l-6 that is what you originally wrote and i think it's great because it: 1) connects W and L in terms of the 280 area; and 2) connects W and L in terms of how they might be equal/ form a square
IOW, go with your original answer, just don't mix up width and length
alright, thank you both so much! now can you tell me how to give you metals? lol
okie dokie!! Thank you both <3