Opti
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Post by Opti on Mar 13, 2015 12:22:43 GMT -5
I feel like such a dumbass. I took the arithmetic section of the posted test and got 18/20. One was an obvious mistake but I can't work out how to get the correct answer on the other one. It's driving me crazy! Can someone please solve this (and show their work)?
Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric to make a curtain, how much will she have left?
a. 1/6 yd b. 1/3 yd c. 1 1/16 yd d. 2 1/3 yd.
The answer is C but I keep getting D and can't work out what I'm doing wrong with the formula. I don't think I'll even attempt the algebra section now . Answer would be C - 1 and 1/6 yard. Multiply by what is left not what is used.
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973beachbum
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Post by 973beachbum on Mar 13, 2015 12:24:16 GMT -5
I thought you said your fallback if the store didn't pan out was to go back to the 6-figure IT stuff, in a different job that you won't hate so much? Personally, I'd focus on that. You're in silicon valley. Tech jobs should be plentiful. They are in my much more buttf*cky, midwestern city. Explaining away a 2 year gap due to trying to start your own biz is perfectly valid and shows ambition. It's not like you sat on your ass during that time. And if you're worried about obsolescence, take some refresher courses in the latest technology. The lapsing clearance shouldn't be that big of an issue. The fact that you have/had a recent one at least tells an employer that you're clearable. Hell, I didn't have one at all when interviewing for my current job. They had to sponsor me, and pay the price to get me one. Getting a degree in something different means you'll have to incur the direct and opportunity costs of that, do all of the work, and then end up starting out at the bottom with the rest of the new grads (with commensurate salary) when you're done. And that's if all goes WELL. Personally, I wouldn't bother unless you REALLY don't want to go back into IT, are willing to pay the (hefty) price to do something different, and if your new degree will be something that's almost guaranteed to show you the money. But that's just me. lots of good thoughts and info here..... Where is bob ross?!?!!? What have you done with him?!?!?!?! Bob has always been smart he just likes to show off his smart ass more.
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gs11rmb
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Post by gs11rmb on Mar 13, 2015 12:27:49 GMT -5
Well at least I was honest and admitted to being a dumbass! My reading comprehension skills also appear to be lacking .
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HoneyBBQ
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Post by HoneyBBQ on Mar 13, 2015 12:28:42 GMT -5
has left = 7/2 x 1/3 (since she uses 2/3) = 7/6 = 1 1/6. (not 1 1/16)
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Deleted
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Post by Deleted on Mar 13, 2015 12:40:12 GMT -5
I feel like such a dumbass. I took the arithmetic section of the posted test and got 18/20. One was an obvious mistake but I can't work out how to get the correct answer on the other one. It's driving me crazy! Can someone please solve this (and show their work)?
Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric to make a curtain, how much will she have left?
a. 1/6 yd b. 1/3 yd c. 1 1/16 yd d. 2 1/3 yd.
The answer is C but I keep getting D and can't work out what I'm doing wrong with the formula. I don't think I'll even attempt the algebra section now . she used 2/3 (which is 2 1/3) and has 1/3 (1 1/16) left over.
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973beachbum
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Post by 973beachbum on Mar 13, 2015 13:08:33 GMT -5
I feel like such a dumbass. I took the arithmetic section of the posted test and got 18/20. One was an obvious mistake but I can't work out how to get the correct answer on the other one. It's driving me crazy! Can someone please solve this (and show their work)?
Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric to make a curtain, how much will she have left?
a. 1/6 yd b. 1/3 yd c. 1 1/16 yd d. 2 1/3 yd.
The answer is C but I keep getting D and can't work out what I'm doing wrong with the formula. I don't think I'll even attempt the algebra section now . she used 2/3 (which is 2 1/3) and has 1/3 (1 1/16) left over. 1/3 of 3 is obviously 1, but 1/3 of 1/2 isn't 1/16th. it is 1/6th. Although I'm guessing that was a typo.
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gs11rmb
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Post by gs11rmb on Mar 13, 2015 13:43:43 GMT -5
Well at least I was honest and admitted to being a dumbass! My reading comprehension skills also appear to be lacking . construct-irrelevant variance. It's the test at fault, not you! That's very kind of you but I just attempted the elementary algebra section and scored 9/20 - I'm going to go out on a limb here and identify myself as the problem . I will try to redeem myself and explain that I correctly answered the ones for which I could remember the rules and formulas!
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Virgil Showlion
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Post by Virgil Showlion on Mar 13, 2015 16:35:08 GMT -5
I feel like such a dumbass. I took the arithmetic section of the posted test and got 18/20. One was an obvious mistake but I can't work out how to get the correct answer on the other one. It's driving me crazy! Can someone please solve this (and show their work)?
Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric to make a curtain, how much will she have left?
a. 1/6 yd b. 1/3 yd c. 1 1/16 yd d. 2 1/3 yd.
The answer is C but I keep getting D and can't work out what I'm doing wrong with the formula. I don't think I'll even attempt the algebra section now . Recall that the von Neumann polyradical of 3 1/2 is subjunct to the eigenmodes of (3x - 1)/2, and the 2nd and 3rd order cumulants are both degenerate, hence by the Cramer-Rao theorem, 2/3 is cohomologous with the compact Morse form of the discriminant (in this case 3/14 + 2x<omega>), and by Weyl's second lemma, coh( d<omega> x py p ) < dim adj A(3) + dim adj A(1/2), allowing us to "pinch" the manifold between the two bounds (20/16 - 2i, 45 + 17i) and iterate (in the Hamilton-Jacobi sense) to reduce the set to one with Lebesgue measure zero. This yields our only candidate solution: 1 1/6 yard. Q.E.D.
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Lizard Queen
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Post by Lizard Queen on Mar 13, 2015 16:50:50 GMT -5
Huh? Feeling dumb here. (1-2/3)*7/2
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Phoenix84
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Post by Phoenix84 on Mar 13, 2015 17:47:01 GMT -5
That's part of the problem I ran into with grad school, it didn't take long for my knowledge of math to start fading after college.
I still feel I have a pretty good grasp on algebra and some geometry, but that's about it at this point.
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TheHaitian
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Post by TheHaitian on Mar 13, 2015 18:26:17 GMT -5
Just passing by and saying good luck with the degree!
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NastyWoman
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Post by NastyWoman on Mar 13, 2015 18:45:39 GMT -5
I feel like such a dumbass. I took the arithmetic section of the posted test and got 18/20. One was an obvious mistake but I can't work out how to get the correct answer on the other one. It's driving me crazy! Can someone please solve this (and show their work)?
Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric to make a curtain, how much will she have left?
a. 1/6 yd b. 1/3 yd c. 1 1/16 yd d. 2 1/3 yd.
The answer is C but I keep getting D and can't work out what I'm doing wrong with the formula. I don't think I'll even attempt the algebra section now . Recall that the von Neumann polyradical of 3 1/2 is subjunct to the eigenmodes of (3x - 1)/2, and the 2nd and 3rd order cumulants are both degenerate, hence by the Cramer-Rao theorem, 2/3 is cohomologous with the compact Morse form of the discriminant (in this case 3/14 + 2x<omega>), and by Weyl's second lemma, coh( d<omega> x py p ) < dim adj A(3) + dim adj A(1/2), allowing us to "pinch" the manifold between the two bounds (20/16 - 2i, 45 + 17i) and iterate (in the Hamilton-Jacobi sense) to reduce the set to one with Lebesgue measure zero. This yields our only candidate solution: 1 1/6 yard. Q.E.D. Not to be difficult but Virgil, you would have not been able to answer this question either since it lacks a "none of the above" option and last time I looked 1 1/6yd was not identical to 1 1/16 yd
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Virgil Showlion
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Post by Virgil Showlion on Mar 13, 2015 19:24:02 GMT -5
Yeah... totally.. or you can just do 7/2*1/3=7/6=1 1/6 Or mentally. A third of three is one. A third of a half is a whole cut into six parts so one sixth. The answer is 1 and 1/6th. Don't confuse her. I've given her all the necessary steps. She doesn't need you mucking it up with your "a third of this is this", "half of that is that" and other bafflegab.
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Opti
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Post by Opti on Mar 13, 2015 19:26:39 GMT -5
Yeah... totally.. or you can just do 7/2*1/3=7/6=1 1/6 Or mentally. A third of three is one. A third of a half is a whole cut into six parts so one sixth. The answer is 1 and 1/6th. Don't confuse her. I've given her all the necessary steps. She doesn't need you mucking it up with your "a third of this is this", "half of that is that" and other bafflegab. Exactly. All she needs is an engine manifold to pinch Virgil with, some EE freezer space...
Oops, I was supposed to be talking about math and snow leopards?
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Virgil Showlion
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Post by Virgil Showlion on Mar 13, 2015 19:29:17 GMT -5
Recall that the von Neumann polyradical of 3 1/2 is subjunct to the eigenmodes of (3x - 1)/2, and the 2nd and 3rd order cumulants are both degenerate, hence by the Cramer-Rao theorem, 2/3 is cohomologous with the compact Morse form of the discriminant (in this case 3/14 + 2x<omega>), and by Weyl's second lemma, coh( d<omega> x py p ) < dim adj A(3) + dim adj A(1/2), allowing us to "pinch" the manifold between the two bounds (20/16 - 2i, 45 + 17i) and iterate (in the Hamilton-Jacobi sense) to reduce the set to one with Lebesgue measure zero. This yields our only candidate solution: 1 1/6 yard. Q.E.D. Not to be difficult but Virgil, you would have not been able to answer this question either since it lacks a "none of the above" option and last time I looked 1 1/6yd was not identical to 1 1/16 yd Sure it is. 1/16 - 1/6 = (1 - 1)/(16 - 6) = 0/10 = 0 Thus the two are equal. D'oyyyy.
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Opti
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Post by Opti on Mar 13, 2015 19:38:15 GMT -5
Snow leopard attempting to do Virgil math. ...between the two bounds ... Hamilton - Jacobi hmmm reduce the set ... does that look like 1/16 or 1/6 tail ... wait I think he says its the same ... (Thank God I'm a snow leopard and don't have to save the dots!)
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Virgil Showlion
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Post by Virgil Showlion on Mar 14, 2015 0:56:36 GMT -5
Snow leopard attempting to do Virgil math. ...between the two bounds ... Hamilton - Jacobi hmmm reduce the set ... does that look like 1/16 or 1/6 tail ... wait I think he says its the same ... (Thank God I'm a snow leopard and don't have to save the dots!)
This is what happens when one attempts to solve the fractions problem using Dark's technique instead of mine.
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8 Bit WWBG
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Post by 8 Bit WWBG on Mar 14, 2015 7:57:57 GMT -5
...:::"The answer is C but I keep getting D and can't work out what I'm doing wrong with the formula. I don't think I'll even attempt the algebra section now":::... You can eliminate D right off the bat because you know that if Erica uses 2/3 of her material, she must have less than half left.   would be more than half. Which means that Erica is a follower of Dave Ramsey because DR is full of cases where people do mathematically impossible things like pay off $40k of debt in 18 months on a $20k salary. Either that, or Erica is the messiah, who can totally use 2/3 of her material but collect 6 baskets of leftovers. ...:::"This is what happens when one attempts to solve the fractions problem using Dark's technique instead of mine.":::... Depending on how his cramming went, this exam may have Dark assuming a similar position as that snow leopard!
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tallguy
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Post by tallguy on Mar 14, 2015 12:28:58 GMT -5
Yeah, yeah, yeah.
Good thing I don't care about a super like from Virgil, no matter how rare it is.
Your lips say "no", but your raising the subject ex post facto says "It should be me getting that super-like. " Let it be known that Tall was technically the first one to correctly solve the problem. No, really, it is not any personal need for affirmation. I have never needed that from anybody. It is instead the somewhat perverse enjoyment one gets from forcing a super like and an acknowledgment of genius for someone who so often disagrees with you.
Not showing the actual solution I guess means that I do not require even that as much as one would think, or at least that it does not outweigh my desire to not preclude others from the enjoyment and sense of accomplishment from solving it on their own. I really am quite a guy!
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Post by Deleted on Mar 14, 2015 13:41:50 GMT -5
I'm guessing a few Khan sessions will have you ready to skip Algebra
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Post by Deleted on Mar 14, 2015 14:34:26 GMT -5
No, absolute value is positive. So that is 8.
The first one is A. Solving works.
Also (x + 1) (x - 3) ... Take first x times x and then times -3, then the 1 times x and then times -3 = x squared - 3x + x -3, which what you started with...
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Deleted
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Post by Deleted on Mar 14, 2015 14:36:45 GMT -5
I don't know how much i'd remember if i was not going through it all again... And again. I know I had forgotten all about imaginary numbers
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mcsangel2
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Post by mcsangel2 on Mar 14, 2015 14:54:05 GMT -5
x to the second power - 2x -3 <--- can't do x2 with the exponent written the way it was on the test A (x+1)(x-3)B (x+1)(x+3) C (x-1)(x+1) D (x+3)(x-1) The correct answer is A, I think, but I have no idea how you're supposed to get there. I know that if you assume x=4 the solution to the equation is 5 and the solution to A is also 5. If you check your work by assuming x=5 the solution to the equation is 12 and the solution to A is also 12. So I'm pretty sure A is correct, but I don't know how you're supposed to get the equation to A without actually solving for X, which is the correct way to do the problem. So I know how to do basic math, but I have no idea how to do actual algebra, even though I'd get this question right on an algebra test. x^2-2x-3 is a quadratic equation. It has three terms. The first is x^2. The second is -2x. The third is -3. To solve, you multiply the coefficient (the number in *front* of the x) of the first and third terms. The first coefficient is 1 (1x^2 is the same thing as x^2), the third is -3, multiply them and you get -3.
Then, you have to list all of the pairs of numbers that you can multiply together to get -3. It's a prime number, so your choices are either -1 and 3, or 1 and -3.
Choose the pair that, when added together, equal the amount of the coefficient of the second term (which is -2). -1 and 3 give you 2, so that's not it. Your answer is 1 and -3, or A.
ETA: edited to correct terminology.
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mcsangel2
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Post by mcsangel2 on Mar 14, 2015 14:55:18 GMT -5
Absolute... duh. I was thinking number set. So I'm like a number set that consists only of the value -8 is equal to the stated value -8. Number sets are in the curly brackets? Number sets are in {} or [], depending on what kind of problem it is.
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Virgil Showlion
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Post by Virgil Showlion on Mar 14, 2015 15:00:04 GMT -5
|-8| is called the "vice" of -8. It's the number you get by "squeezing" all the symbols inside the brackets together into a single symbol (like a vice).
In this case, the vice of -8 is 8 because the - just gets squeezed inside the middle part of the 8.
|13| = 8 because the 1 and the 3 squeeze together to make an 8.
|71Γ| = T because the three stems all squeeze together to form the stem of the T.
You get the idea.
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Virgil Showlion
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Post by Virgil Showlion on Mar 14, 2015 15:18:14 GMT -5
x to the second power - 2x -3 <--- can't do x2 with the exponent written the way it was on the test A (x+1)(x-3)B (x+1)(x+3) C (x-1)(x+1) D (x+3)(x-1) The correct answer is A, I think, but I have no idea how you're supposed to get there. I know that if you assume x=4 the solution to the equation is 5 and the solution to A is also 5. If you check your work by assuming x=5 the solution to the equation is 12 and the solution to A is also 12. So I'm pretty sure A is correct, but I don't know how you're supposed to get the equation to A without actually solving for X, which is the correct way to do the problem. So I know how to do basic math, but I have no idea how to do actual algebra, even though I'd get this question right on an algebra test. x^2-2x-3 is a quadratic equation. It has three terms. The first is x^2. The second is -2x. The third is -3. To solve, you multiply the exponent (the number in *front* of the x) of the first and third terms. The first exponent is 1 (1x^2 is the same thing as x^2), the third is -3, multiply them and you get -3.
Then, you have to list all of the pairs of numbers that you can multiply together to get -3. It's a prime number, so your choices are either -1 and 3, or 1 and -3.
Choose the pair that, when added together, equal the amount of the exponent of the second term (which is -2). -1 and 3 give you 2, so that's not it. Your answer is 1 and -3, or A.
In all seriousness: the leading numbers are called "coefficients" ("exponents" are the superscripted numbers), and you should be mindful of the fact that the above procedure doesn't work if the x 2 coefficient isn't 1 (or more generally, if the roots aren't integers).
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mcsangel2
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Post by mcsangel2 on Mar 14, 2015 15:48:11 GMT -5
Sum Dum Gai, I was in your position a couple of years ago. I graduated high school in 1990, attended community college on/off until 1996. My last math course was college algebra in 1996, which is the highest math needed for a liberal arts diploma. However, I did not finish college and did not even finish an associate's degree.
I went back to school in 2010, now to finish a degree in accounting in which I have been employed since 1996. This is a business degree, which requires brief calculus, business calculus, and statistics on top of college algebra. The requirements for cc's in Arizona are similar to yours, in that you must have finished either the previous math course or taken a placement test with the appropriate score within the last 2 years. I didn't even try to take it, like you did today, I just did practice ones online. I forgot EVERYTHING.
I did not want to take all those courses over again, so I decided to hire a tutor. He was a retired math teacher, found him on Wyzant.com. There are lots of other sites where you can find tutors, too.
This was such a good decision. I had to start all over with high school freshman algebra. But that only took a month (meeting once a week), algebra II took 2 months, and then the next 4 months was an overall brushup of algebra/basic trig. So I met with the tutor for 7 months, while I was taking other classes I still needed, and then took a placement test in January 2014. I took the highest math placement test they had available, and I scored on the low side, but high enough to enroll in any math class I wanted. (I could have enrolled into calc I, but I don't need a class that high).
I will say that the test scoring was conservative relative to how comfortable I felt in the class, which was very. I just needed the review, everything was fine. I was NOT a strong math student when I was younger, never having made anything higher than B including in college algebra. I made As in brief calculus and business calculus. I'm halfway through statistics which is not going as smoothly, but I still think I'll probably get a B, and this is the very last class I need to get my associate of business.
Get a tutor. You will lose your mind if you sit through entire semesters of algebra classes that you don't need to catch up. You just need to review. You can also save time that way while doing it while taking other classes.
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mcsangel2
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Post by mcsangel2 on Mar 14, 2015 15:54:23 GMT -5
x^2-2x-3 is a quadratic equation. It has three terms. The first is x^2. The second is -2x. The third is -3. To solve, you multiply the exponent (the number in *front* of the x) of the first and third terms. The first exponent is 1 (1x^2 is the same thing as x^2), the third is -3, multiply them and you get -3.
Then, you have to list all of the pairs of numbers that you can multiply together to get -3. It's a prime number, so your choices are either -1 and 3, or 1 and -3.
Choose the pair that, when added together, equal the amount of the exponent of the second term (which is -2). -1 and 3 give you 2, so that's not it. Your answer is 1 and -3, or A.
In all seriousness: the leading numbers are called "coefficients" ("exponents" are the superscripted numbers), and you should be mindful of the fact that the above procedure doesn't work if the x2 coefficient isn't 1 (or more generally, if the roots aren't integers). Oops, you are right! My bad. You can also use the quadratic formula, but it's very long and annoying.
Are you sure the first procedure doesn't work if the first coefficient isn't 1? I know that *sometimes* it doesn't work, and you need to use the quadratic formula, but I thought sometimes it did.
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weltschmerz
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Post by weltschmerz on Mar 14, 2015 16:34:13 GMT -5
I absolutely suck at math but I'd convert the whole thing to inches and take it from there. She bought 126 inches of fabric. What's two thirds of 126? She used 84 inches of fabric to make the curtain. She has 42 inches of fabric left. Pretty easy from there.
Oh, screw that!!
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teen persuasion
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Post by teen persuasion on Mar 14, 2015 18:34:26 GMT -5
|-8| is called the " vice" of -8. It's the number you get by "squeezing" all the symbols inside the brackets together into a single symbol (like a vice). In this case, the vice of -8 is 8 because the - just gets squeezed inside the middle part of the 8. |13| = 8 because the 1 and the 3 squeeze together to make an 8. |71Γ| = T because the three stems all squeeze together to form the stem of the T. You get the idea. And now we've crossed over into the playing with words thread. I kept tripping over the "c" in vice, trying to figure out why you were trying to squeeze with a moral failing rather than a tool. I didn't realize that American English distinguishes between the words with a different spelling, and British/Canadian English does not. "In American English, a vice is an immoral habit or practice, while a vise is a tool with closable jaws for clamping things. But in British English, the tool is spelled like the sin: vice." (Bryce A. Garner, Garner's Modern American Usage, 3rd ed. Oxford University Press, 2009) I'm accustomed to many of the spelling differences, but this is the first time I've noticed this one.
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