EvoN1020v Posted January 27, 2006 Posted January 27, 2006 From now on, I will use this thread to inquire my math questions that I have. I don't want to create a new thread for everytime when I have a single, small math question. Waste of time in my opinion. Also, to let everybody know, those math questions will never be my homework. They are just some bogus math questions that I see around or on the Internet. IT'S NOT MY HOMEWORK. Awhile ago, I just found this old Fermat Competition booklet in my bedroom that I entered last year (A Canadian National Math Contest), and I have this question. When [math]a=\frac{1}{2}[/math] and [math]b=\frac{2}{3}[/math], what what does [math]\frac{6a + 18b}{12a + 6b}[/math] equals? I found the answer to be [math]\frac{3}{2}[/math] or rather [math]1.5[/math]. I know this is a really simple question, but I was just wondering how I can find the answer quickly? By using fraction reduction, or substution, or anything that can give me the answer quickly, rather than do all the work necessary.
computerages Posted January 28, 2006 Posted January 28, 2006 Use your imaginations. Such as, when you subsitute "a" with 1/2 or "b" with 2/3, do all that in your head without writing anything down. It seems like JHS math problem so you may not even need to use your math skills, but imaginations.
EvoN1020v Posted January 28, 2006 Author Posted January 28, 2006 Yeah I guess you could do it in your head such as [math]18 \cdot \frac{2}{3}[/math] is simply 12. I approached another question that do not have the answers that I found in its multiple choice. If [math](x-4)(5x+2) = 0[/math], then the two possible values of [math]5x + 2[/math] are: (A) -4 and 2/5 (B) 0 and -18 © 0 and 22 (D) 0 and 4 (E) 4 and 22 At first, I thought okay, then I got x to be simply 0 and [math]\frac{-2}{5}[/math], but as I scrolled through the answer, its not there. I was thinking, okay? Next, I used the quadratic formula for the whole equation which comes to be [math]5x^{2} - 18x - 8[/math]. It yielded 4 and [math]\frac{-1}{4}[/math], but it is for the whole equation, not just [math]5x + 2[/math] itself. Again, I looked at the choices, and none are there. Any help?
clarisse Posted January 28, 2006 Posted January 28, 2006 the answer is C because in order for (x-4)(5x+2) = 0, at least one should be equal to zero (because something times something equals zero you see?) and if it did not equal zero, then the other bracket should be equal to zero so x-4=0 and x=4 and you substitute this into 5x+2=0 and the answer is 22
EvoN1020v Posted January 28, 2006 Author Posted January 28, 2006 But if you input the "0" in [math]5x + 2 = 0[/math], therefore [math]5(0) + 2[/math] would be 2?
TD Posted January 28, 2006 Posted January 28, 2006 They're asking for the possible values of "5x+2", not of x. So when you say 0 is a possibility, this doesn't mean "x = 0" but "5x+2 = 0". C is the correct answer.
EvoN1020v Posted January 29, 2006 Author Posted January 29, 2006 OK, I understand abit of what the question is implying. Say, TD, how did you figure out the answer? Can you please explain it clearly.
Connor Posted January 29, 2006 Posted January 29, 2006 well, in order for the equation to =0, 5x+2=0 (first answer) or x-4=0 if x-4=0, x=4 and 5x+2=22 (second answer)
EvoN1020v Posted January 30, 2006 Author Posted January 30, 2006 I have another question: Sam bicycles at 16 km/h and Chris bicycles at 24 km/h. At noon, Sam is 1 km north of Chris, and each begins to ride north. How many minutes will it take for Chris to catch Sam? I have no idea what to do. I assume using displacement formula or something?
cosine Posted January 30, 2006 Posted January 30, 2006 I have another question: Sam bicycles at 16 km/h and Chris bicycles at 24 km/h. At noon' date=' Sam is 1 km north of Chris, and each begins to ride north. How many minutes will it take for Chris to catch Sam? I have no idea what to do. I assume using displacement formula or something?[/quote'] well firstly: distance = speed * time let noon be t = 0 hours then Sam's distace north (S) = 1 + 16*t C = 24t So to see when Chris catched up to Sam, set their distances equal to each other: 1 + 16t = 24t 8t = 1 t = 1/8 So Chris catches up with Sam an eigth of an hour after noon.
EvoN1020v Posted January 30, 2006 Author Posted January 30, 2006 Very clever. I never thought of appointing the 1 km north as 1 unit up on the y-axis. Another question: If 50% of P equals 20% of Q, then P, as a percent of Q, is ? 0.5x = o.2y doesn't seem to fit the question.
Klaynos Posted January 30, 2006 Posted January 30, 2006 0.5x=0.2y (1) 1x=ty (2) Treat them as simultaniouse equations do (1)/(2) and solve for t
Klaynos Posted January 30, 2006 Posted January 30, 2006 What is "simultaniouse equations"? A set of equations that are all true at the same time. If you just do the first equation dived by the second equation you should get one fraction = another fraction with no x's or y's and then then reorder them to get what the variable equals. Or you can do it simply using logical arguments. Which is mathmatically easier but it isn't a proof just a "by inspection I can see the answer is"
EvoN1020v Posted January 31, 2006 Author Posted January 31, 2006 I never thought I could solve for 2 simultaneous equations that have 3 different variables, but it comes to that x is only 1. So I used substution method to discover the answer, not the elimination that Klaynos recommended. I got an answer for t= 0.4. This is the answer for Q. I multipled 0.5 to the answer, and it yielded 0.2 for the value of P. Am I correct?
RedAlert Posted January 31, 2006 Posted January 31, 2006 I never thought I could solve for 2 simultaneous equations that have 3 different variables' date=' but it comes to that x is only 1. So I used substution method to discover the answer, not the elimination that [b']Klaynos[/b] recommended. I got an answer for t= 0.4. This is the answer for Q. I multipled 0.5 to the answer, and it yielded 0.2 for the value of P. Am I correct? Well, this is how I would do it. We need to find p/q (and then times it by a hundred to get the percent). Since 0.5x = 0.2y Then let y = 1. In which case: 0.5x = 0.2 x = 0.2/0.5 x = 0.4 Therefore: 0.5 * 0.4 = 0.2y 0.2 = 0.2y 0.2/0.2 = y 1 = y 0.4/1 = p/q 0.4*100 = p/q 40% = p/q * 100<---- This step is to give us p as a percent of q Therefore, our final answer would be: 40%, or 0.4 as you said. So yes, you are correct. By the way, are you in Grade 10? Because I am, and I am doing the same stuff as you are.
EvoN1020v Posted January 31, 2006 Author Posted January 31, 2006 No. It was a question extracted from an old Fermat Math Competition booklet. I was skeptical at the requestion of the question because English is my second language. "then P, as a percent of Q", is rather confusing to me. I think it means, what is the answer for P, while Q is a percent too. By the way, I am in grade 12, so apparently I have been forgetting grade 10 math, so I'm doing this to refresh my memory
Klaynos Posted January 31, 2006 Posted January 31, 2006 I never thought I could solve for 2 simultaneous equations that have 3 different variables' date=' but it comes to that x is only 1. So I used substution method to discover the answer, not the elimination that [b']Klaynos[/b] recommended. I got an answer for t= 0.4. This is the answer for Q. I multipled 0.5 to the answer, and it yielded 0.2 for the value of P. Am I correct? Yeah that's just anotehr way of solving the simultaniouse equation. I just did it like that because I started off thinking about it as a ratio and worked backwards. I also got 40% to be the answer (0.4)
EvoN1020v Posted January 31, 2006 Author Posted January 31, 2006 Four different numbers a, b, c, and d are chosen from the list -1, -2, -3, -4, and -5. The largest possible value for the expression [math]a^{b} + c^{d}[/math] is ? I already figured out the answer, but wanted to see what you guys got.
Klaynos Posted January 31, 2006 Posted January 31, 2006 What max value do you get? I take it it's probably a low +ve fraction possibly around 1?
Klaynos Posted January 31, 2006 Posted January 31, 2006 Nope. A hint: It's greater than 1. Oh I know that. just cba to work it out
EvoN1020v Posted January 31, 2006 Author Posted January 31, 2006 Stop wasting my time! Do you have any finite answer to my question? If anybody got the same answer as I do, then there's a challenge question next.
EvoN1020v Posted January 31, 2006 Author Posted January 31, 2006 What is the derivative of [math]3x^{3} - 5x^{2} + 2x + 13[/math]? As I know that [math]k \cdot x^{n}[/math] is equal to [math]n \cdot k \cdot x^{n-1}[/math]. Therefore: [math](3 \cdot 3 \cdot x^{3-1})-(2 \cdot 5 \cdot x^{2-1})+(1 \cdot 5 \cdot x^{1-1}) + 13...?[/math] [math]\rightarrow 9x^{2} - 10x + ....?[/math] I have no idea how to get the final value? Because the answer said to be [math]9x^{2} - 10x + 2[/math].
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