Remember when you were in grade 10? I was just looking through my brother’s textbook and realized just how hard an easy thing can be. Right now he’s doing Algebra, making his own equations and solving them. It’s in the Grade 10 Academic Math Ciriculum from Ontario if anyone wants to indulge. Heres the quiz:
A rectangle with a perimeter of 40m is 2m longer than it is wide. Determine the dimensions of the rectangle.
A publishing company in Idaho mails 1040 brochures promoting a new university textbook. Each brochure costs $0.46 to mail within North America and $0.48 to mail elsewhere. If the total mailing cost was $483, how many brochures were mailed to universities in North America.
Tip: You’re only looking for an intersect here, not both coordinates.
Extra Points for this one.
2y = 5x-3
y= 2.5x-3 (those are x’s for variables.)
Whats wrong?
Dopers, you’re getting marked on:
Let statements (let x represent… let y represent…) use any variable you wish.
Equations.
Solving. Using Substitution
Conclusion. X is equal to blah and Y is equal to blah blah.
I’d post the spoilers but I can get that whole black table thing. So I’ll put them here:
9m x 11m
810 Were mailed.
No solution. Everything cancels out therefore they are parallel.
I can do grade 10 algebra. Having the answers in your OP without much warning makes it a little hard to be objective, though.
[spoiler]Anyway, for #1, let x represent the length of the rectangle in meters. Then x - 2 is the width of the rectangle, also in meters. So x satisfies 4x - 4 = 40 (to be absolutely correct, I should mark both the constants as meters, but no one does that, so I won’t). Then 4x = 44, and x = 11. Therefore, the rectangle is 9 m x 11 m.
As for the second one. Let x represent the brochures mailed to North America, and y the brochures mailed elsewhere. .46x + .48y = 483, and x + y = 1040. Applying that ol’ Gauss-Jordan reduction algorithm, we get that x = 810 and y = 230. Therefore, 810 brochures were mailed out to North America.
For the third one, 2y = 5x - 3 is equivalent to y = 2.5x - 1.5. If there is an intercept between the two lines, it’s at a point (x, y) that satisfies y = 2.5x - 3 and y = 2.5x - 1.5. Then by the transitive property of equality, 2.5x - 3 = 2.5x - 1.5, and we can conclude that 1.5 = 0. This is clearly false (although math beyond the 10th grade level is necessary to show this result), so there is no intercept.[/spoiler]
For those who are feeling pretty good about it at this point, here’s a bonus question: what is a variable?
Now… i was revisiting my senior math today (i haven’t used it in 5 years- since i first started university)
who here can remember their differential calculus?
You can do #1 with algebra, but that’s using a cannon to kill a mosquito. A square with perimeter of 40m has four sides of 10m. If it’s a rectangle with long sides 2m longer than the short sides, sides of 11m and 9m kinda falls into your lap as the natural first try in ‘trial and error’.
I interpreted the question in #3 as “we reduced the first equation to the second one, what did we do wrong,” which (under this interpretation) was failure to divide the last term (the constant term of -3) by 2. Which tells you what’s wrong with it as a system of equations.
And ultrafilter has already said all that needs saying about #2.
Depends on your context. In programming a variable is a named area of memory where data is stored. Algebra always did suck when it came to giving your variables descriptive names.
I get -3=0. Either one of us did something wrong or that’s one funky problem:)
I think what you did wrong was forgetting to multiply the 1.5 by 2. However, the last real math I had was in '99 for the calc AB AP exam, so that might be why:)
We’re both right. If -3 = 0, then you can add three to both sides to get 0 = 3. And then you can divide by 2 to get 1.5 = 0. Then you can multiply by a/1.5 to get a = 0, and add c to get a + c = c. So from a false equality, you can conclude that any pair of numbers is equal.
You know, this reminds me of something that seriously ticked me off. There I was, innocently taking the GRE General test (The Graduate Record Exam, which most grad schools require, for the non-US and non-grad-school inclined among us. There is a general test and then several subject-specific tests). And there was a question on multiplying polynomials. I learned that in the 8th freaking grade! Reaching deep into my 22 year-old brain, and with a prayer of gratitude towards Mrs. Chan, my 8th grade math teacher, I did solve the problem. The grad school exam is quizzing me on material that I was taught 8 years ago and has never come up since? I mean, what’s with that?
So, middle school math teachers, when you hear that familiar whine, “why do we have to learn this!?” You can safely answer “because it will be on your GRE.”
I’m not sure. Here’s the ones I remember off the top of my head:
Table of derivatives and integrals
Trig table
Quadratic formula
The t-shirt with the “2=1” proof that DrMatrix finished for me just now.
The one that begins “Integration is as easy as” followed by an equation with, on one side, the sum of the integral of e[sup]x[/sup] from minus-infinity to 0 and the integral of sin(x) from 0 to pi/2, and on the other, the sum of 1/2[sup]n[/sup] as n goes from 0 to infinity. (That’s my favorite.)
A T-shirt one of my grad profs had custom-made for his commutative-rings class, IIRC. It’s got all sorts of funky algebraic terminology on it in various typefaces.
The “I think you should be more explicit in Step 2” T-shirt (“then a miracle occurs”) that everybody and their brother has seen, so I never wear it.
I think that’s polished my nerd credentials sufficiently for now. Anyone care to match math shirt collections with me?
That reminds me of when I used to teach the Chain Rule in Calc I. I’d call the inner function, the one you’re initially treating as a variable, something like Harry or George. I’d explain that the derivative of f(Harry) would be f’(Harry)*(Harry)’. It seemed to get the idea of how to take chain-rule derivatives across about as well as anything I tried.
And then when I’d be explaining integration by substitution later in the semester, I’d pull that explanation out again to answer the inevitable question about where the chain-rule derivative vanished to when you were integrating rather than differentiating.
