Casper Posted February 28, 2011 Report Share Posted February 28, 2011 Find the standard form of the equation of the ellipse: 64x^2 + 49y^2 – 768x – 686y + 1569 = 0Bonus for finding the center, vertices, foci, and eccentricity of the ellipse.Enjoy. Quote Link to comment Share on other sites More sharing options...
Bad324 Posted February 28, 2011 Report Share Posted February 28, 2011 Find the standard form of the eclipse:BOOM! what do i win? Quote Link to comment Share on other sites More sharing options...
flounder Posted February 28, 2011 Report Share Posted February 28, 2011 So I take it Casper is taking a math class at the present time? Quote Link to comment Share on other sites More sharing options...
jblosser Posted February 28, 2011 Report Share Posted February 28, 2011 the answer is 3actually, the answer is "hell if I know". Quote Link to comment Share on other sites More sharing options...
Casper Posted February 28, 2011 Author Report Share Posted February 28, 2011 So I take it Casper is taking a math class at the present time?Yup. And a logical thinking class. That one is a riot. I just keep thinking, "duh". And somehow the average grade in the class is an 84%? How? I have a 98.7%. The only points I missed were on one assignment I miscopied the questions. I missed the first one. Quote Link to comment Share on other sites More sharing options...
John Posted February 28, 2011 Report Share Posted February 28, 2011 If a brick counter balances three quarters of a pound plus three quarters of a brick, how many pounds does the whole brick weigh? Quote Link to comment Share on other sites More sharing options...
jblosser Posted February 28, 2011 Report Share Posted February 28, 2011 like i said before, the answer is 3.jeesh. Quote Link to comment Share on other sites More sharing options...
CoolWhip Posted February 28, 2011 Report Share Posted February 28, 2011 Wait, isn't the answer always supposed to be 42? Quote Link to comment Share on other sites More sharing options...
Casper Posted February 28, 2011 Author Report Share Posted February 28, 2011 If a brick counter balances three quarters of a pound plus three quarters of a brick, how many pounds does the whole brick weigh?3lbs? Quote Link to comment Share on other sites More sharing options...
gen3flygirl Posted February 28, 2011 Report Share Posted February 28, 2011 Have you tried Googling it? Quote Link to comment Share on other sites More sharing options...
Beegreenstrings Posted February 28, 2011 Report Share Posted February 28, 2011 the answer is always C. When in doubt C! Quote Link to comment Share on other sites More sharing options...
jblosser Posted February 28, 2011 Report Share Posted February 28, 2011 like i said before, the answer is 3.jeesh.3lbs?copy cat.The answer is Herbert Hoover. Quote Link to comment Share on other sites More sharing options...
Disclaimer Posted February 28, 2011 Report Share Posted February 28, 2011 Find the standard form of the equation of the ellipse: 64x^2 + 49y^2 – 768x – 686y + 1569 = 0Bonus for finding the center, vertices, foci, and eccentricity of the ellipse.Enjoy. Std form -> (x-h)^2/a^2 + (y-k)^2/b^2 = 164x^2 + 49y^2 – 768x – 686y + 1569 = 064x^2 – 768x + 2304 + 49y^2 – 686y + 2401-3136 = 064(x^2-12x+36)+49(y^2-14y+49)-3136 = 064(x-6)^2+49(y-7)^2=3136(x-6)^2/7^2+(y-7)^2/8^2=1Center @ (h, k): (6,7)Vertices @ +/- y where x=6:y^2-14y+49=64y^2-14y-15=0(y+1)(y-15)=0y = (-1 and 15); therefore vertices @ (6,-1) & (6,15) for the major axisVertices @ +/-x where y=7:x^2-12x+36=49x^2-12x-13=0(x-13)(x+1)=0x = (-1 and 13); therefore vertices @ (7,-1) & (7,13) for the minor axisfocii (a^2-b^2=c^2):64-49=c^215=c^2; therefore focii are (6, 7+SQRT(15)) & (6,7-SQRT(15))Eccentricity = SQRT(15)/8How many are wrong this time? I think I double-checked it all...? Quote Link to comment Share on other sites More sharing options...
JStump Posted February 28, 2011 Report Share Posted February 28, 2011 Std form -> (x-h)^2/a^2 + (y-k)^2/b^2 = 164x^2 + 49y^2 – 768x – 686y + 1569 = 064x^2 – 768x + 2304 + 49y^2 – 686y + 2401-3136 = 064(x^2-12x+36)+49(y^2-14y+49)-3136 = 064(x-6)^2+49(y-7)^2=3136(x-6)^2/7^2+(y-7)^2/8^2=1Center @ (h, k): (6,7)Vertices @ +/- y where x=6:y^2-14y+49=64y^2-14y-15=0(y+1)(y-15)=0y = (-1 and 15); therefore vertices @ (6,-1) & (6,15) for the major axisVertices @ +/-x where y=7:x^2-12x+36=49x^2-12x-13=0(x-13)(x+1)=0x = (-1 and 13); therefore vertices @ (7,-1) & (7,13) for the minor axisfocii (a^2-b^2=c^2):64-49=c^215=c^2; therefore focii are (6, 7+SQRT(15)) & (6,7-SQRT(15))Eccentricity = SQRT(15)/8How many are wrong this time? I think I double-checked it all...?This shit is why I'm not an engineering major anymore lol Quote Link to comment Share on other sites More sharing options...
Casper Posted February 28, 2011 Author Report Share Posted February 28, 2011 Std form -> (x-h)^2/a^2 + (y-k)^2/b^2 = 164x^2 + 49y^2 – 768x – 686y + 1569 = 064x^2 – 768x + 2304 + 49y^2 – 686y + 2401-3136 = 064(x^2-12x+36)+49(y^2-14y+49)-3136 = 064(x-6)^2+49(y-7)^2=3136(x-6)^2/7^2+(y-7)^2/8^2=1Center @ (h, k): (6,7)Vertices @ +/- y where x=6:y^2-14y+49=64y^2-14y-15=0(y+1)(y-15)=0y = (-1 and 15); therefore vertices @ (6,-1) & (6,15) for the major axisVertices @ +/-x where y=7:x^2-12x+36=49x^2-12x-13=0(x-13)(x+1)=0x = (-1 and 13); therefore vertices @ (7,-1) & (7,13) for the minor axisfocii (a^2-b^2=c^2):64-49=c^215=c^2; therefore focii are (6, 7+SQRT(15)) & (6,7-SQRT(15))Eccentricity = SQRT(15)/8How many are wrong this time? I think I double-checked it all...?Only thing is you didn't simplify 7^2 (49) and 8^2 (64). Plus you have to be a bit more careful with those brackets. You have exponents currently being divided and whatnot. The correct answer for full credit is:((x-6)^2)/49 + ((y-7)^2)/64 = 1But I'd only mark off a point or two for that. Quote Link to comment Share on other sites More sharing options...
Disclaimer Posted February 28, 2011 Report Share Posted February 28, 2011 I put it in std form... which is a^2 and b^2. So I'd tell my math prof to lick my taint if he took points off for that since std form was with the denominators squared (because it's easier to use them later to find other attributes). Quote Link to comment Share on other sites More sharing options...
Casper Posted February 28, 2011 Author Report Share Posted February 28, 2011 I put it in std form... which is a^2 and b^2. So I'd tell my math prof to lick my taint if he took points off for that since std form was with the denominators squared (because it's easier to use them later to find other attributes).Nada amigo. You still need to simplify. Just take for example (a^2)+(b^2). If you know a is 2 and b is 5, you don't answer with (2^2)+(5^2). You answer with 29. That's a basic foundation of equations. Always, always, always simplify. We're talking 4th grade math here dude. Always simplify. Quote Link to comment Share on other sites More sharing options...
Disclaimer Posted February 28, 2011 Report Share Posted February 28, 2011 That still wouldn't change the fact I'd tell them to lick my taint. Quote Link to comment Share on other sites More sharing options...
Casper Posted February 28, 2011 Author Report Share Posted February 28, 2011 That still wouldn't change the fact I'd tell them to lick my taint.If I were your prof and I only took a point or two off, then you told me to lick your taint even though you're obviously wrong, I'd kick you out of the class. I'd probably be a dick about too, and file an academic ethics complaint with your dean. Quote Link to comment Share on other sites More sharing options...
TSB67 Posted February 28, 2011 Report Share Posted February 28, 2011 Plus you have to be a bit more careful with those brackets. You have exponents currently being divided and whatnot. Those exponents are not divided in the standrad PEMDAS order of operations. You the student or the teacher? 1 Quote Link to comment Share on other sites More sharing options...
Casper Posted February 28, 2011 Author Report Share Posted February 28, 2011 Those exponents are not divided in the standrad PEMDAS order of operations. You the student or the teacher?My point was this: (x-6)^2/7^2+(y-7)^2/8^2=1Could be mistaken for: (x-6)^(2/7^(2+(y-7)^(2/8^(2))))=1Break it down: (x-6)^2/7That in itself could be mistaken for the quantity of x-6 to the 2/7th power. I agree with you though, following the order of operations it should be fine. Hence I wouldn't take off points for it. I'd only take off points for not fully simplifying. But I would note to be careful about things like that. Quote Link to comment Share on other sites More sharing options...
TSB67 Posted February 28, 2011 Report Share Posted February 28, 2011 Break it down: (x-6)^2/7That in itself could be mistaken for the quantity of x-6 to the 2/7th power. I respectfully disagree.If I wanted to be a prick, I would take points off for you cluttering the thing up with unnecessary parentheses while showing a lack of understanding of the standard order of operations. Quote Link to comment Share on other sites More sharing options...
Casper Posted February 28, 2011 Author Report Share Posted February 28, 2011 I respectfully disagree.If I wanted to be a prick, I would take points off for you cluttering the thing up with unnecessary parentheses while showing a lack of understanding of the standard order of operations. In all honesty, if you typed it out like that on a paper I probably wouldn't give any credit since you didn't use tools provided within the class required text editing software. Quote Link to comment Share on other sites More sharing options...
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