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Another math trivia question


Casper

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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.

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Find the standard form of the equation of the ellipse:

64x^2 + 49y^2 – 768x – 686y + 1569 = 0

Bonus for finding the center, vertices, foci, and eccentricity of the ellipse.

Enjoy. :D

Std form -> (x-h)^2/a^2 + (y-k)^2/b^2 = 1

64x^2 + 49y^2 – 768x – 686y + 1569 = 0

64x^2 – 768x + 2304 + 49y^2 – 686y + 2401-3136 = 0

64(x^2-12x+36)+49(y^2-14y+49)-3136 = 0

64(x-6)^2+49(y-7)^2=3136

(x-6)^2/7^2+(y-7)^2/8^2=1

Center @ (h, k): (6,7)

Vertices @ +/- y where x=6:

y^2-14y+49=64

y^2-14y-15=0

(y+1)(y-15)=0

y = (-1 and 15); therefore vertices @ (6,-1) & (6,15) for the major axis

Vertices @ +/-x where y=7:

x^2-12x+36=49

x^2-12x-13=0

(x-13)(x+1)=0

x = (-1 and 13); therefore vertices @ (7,-1) & (7,13) for the minor axis

focii (a^2-b^2=c^2):

64-49=c^2

15=c^2; therefore focii are (6, 7+SQRT(15)) & (6,7-SQRT(15))

Eccentricity = SQRT(15)/8

How many are wrong this time? I think I double-checked it all...?

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Std form -> (x-h)^2/a^2 + (y-k)^2/b^2 = 1

64x^2 + 49y^2 – 768x – 686y + 1569 = 0

64x^2 – 768x + 2304 + 49y^2 – 686y + 2401-3136 = 0

64(x^2-12x+36)+49(y^2-14y+49)-3136 = 0

64(x-6)^2+49(y-7)^2=3136

(x-6)^2/7^2+(y-7)^2/8^2=1

Center @ (h, k): (6,7)

Vertices @ +/- y where x=6:

y^2-14y+49=64

y^2-14y-15=0

(y+1)(y-15)=0

y = (-1 and 15); therefore vertices @ (6,-1) & (6,15) for the major axis

Vertices @ +/-x where y=7:

x^2-12x+36=49

x^2-12x-13=0

(x-13)(x+1)=0

x = (-1 and 13); therefore vertices @ (7,-1) & (7,13) for the minor axis

focii (a^2-b^2=c^2):

64-49=c^2

15=c^2; therefore focii are (6, 7+SQRT(15)) & (6,7-SQRT(15))

Eccentricity = SQRT(15)/8

How many are wrong this time? I think I double-checked it all...?

This shit is why I'm not an engineering major anymore lol

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Std form -> (x-h)^2/a^2 + (y-k)^2/b^2 = 1

64x^2 + 49y^2 – 768x – 686y + 1569 = 0

64x^2 – 768x + 2304 + 49y^2 – 686y + 2401-3136 = 0

64(x^2-12x+36)+49(y^2-14y+49)-3136 = 0

64(x-6)^2+49(y-7)^2=3136

(x-6)^2/7^2+(y-7)^2/8^2=1

Center @ (h, k): (6,7)

Vertices @ +/- y where x=6:

y^2-14y+49=64

y^2-14y-15=0

(y+1)(y-15)=0

y = (-1 and 15); therefore vertices @ (6,-1) & (6,15) for the major axis

Vertices @ +/-x where y=7:

x^2-12x+36=49

x^2-12x-13=0

(x-13)(x+1)=0

x = (-1 and 13); therefore vertices @ (7,-1) & (7,13) for the minor axis

focii (a^2-b^2=c^2):

64-49=c^2

15=c^2; therefore focii are (6, 7+SQRT(15)) & (6,7-SQRT(15))

Eccentricity = SQRT(15)/8

How 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 = 1

But I'd only mark off a point or two for that.

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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.

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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. :nono:

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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=1

Could be mistaken for: (x-6)^(2/7^(2+(y-7)^(2/8^(2))))=1

Break it down: (x-6)^2/7

That 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.

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Break it down: (x-6)^2/7

That 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. :D

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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. :D

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. :D

:lol:

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