+2446 To solve this problem, we must figure out how much square feet the entire room is. We know the length and the width of the room. Since, this room is rectangular, we can use that information to figure out how much square feet this room covers.
| \(8\frac{3}{4}*9\frac{1}{3}\) | First, convert both fractions to improper fractions so that we can multiply them. I will convert each one separately. |
| \(8\frac{3}{4}=\frac{4*8+3}{4}=\frac{35}{4}\) | |
| \(9\frac{1}{3}=\frac{3*9+1}{3}=\frac{28}{3}\) | Now, let's multiply both together. |
| \(\frac{35}{4}*\frac{28}{3}\) | 4 and 28 have a common factor of 4. Doing this simplification makes it easier computationally. |
| \(\frac{35}{1}*\frac{7}{3}\) | Now, do the multiplication. One-digit multiplication is simpler than doing 2-digit. |
| \(\frac{245}{3}\) | |
We aren't done yet, though! The question is asking how much carpet is remaining after covering the entire floor. This requires the subtraction operator.
| \(85-\frac{245}{3}\) | Make 85 a fraction in which it has a denominator of 3. |
| \(\frac{85}{1}*\frac{3}{3}=\frac{255}{3}\) | |
| \(\frac{255}{3}-\frac{245}{3}\) | Now, we can subtract knowing that we have common denominators. |
| \(\frac{10}{3}ft^2\) | And of course, do not forget to include units, if applicable. |
Therefore, after completely covering the \(\left(8\frac{3}{4}\right)^{'} * \left(9\frac{1}{3}\right)^{'}\)rectangular room in carpet, Rodger will have \(\frac{10}{3}ft^2=3\frac{1}{3}ft^2=3.\overline{33}ft^2\) of carpet left.
+2236 Guest all you gave was an example of arrogance squared.
Sadly, melody's way of answering the questions is clumsy and complicated, and takes a lot of time when the numbers get larger (assuming you remember to count every combinations exactly once).
This isn’t sad, nor is this clumsy. It may be unwieldy because of its size, but it’s not clumsy. Complicated? Yes, of course, it is, and no matter how it’s answered it’s a complicated question. Melody’s presentation explains what is going on inside the formulas and algorithms used to solve this. Very un-advanced students are the target audience, here. They don’t know what generating function is. I doubt there are five members that know what a generating function is –let alone how do one.
This question is a famous one in combinatorics and has a much, MUCH simpler answer than the answers you gave. after reading Jz1234's last post i see he ALMOST got the right answer (and that is VERY impressive, well done!).
No, he didn’t almost get the right answer.
jz1234's answers works only if we dont mention any upper bound in the question.
No it doesn’t. If the goal is Mars, that formula doesn’t get you the moon.
Let me explain:
[Sadly] Your explanation is laborious [Clumsy] and doesn’t produce a very usable result.
As far as i know there is no general, nice formula that finds the number of rolls, but that is the best way i can think of for solving this question
Well you don’t know much! I know of a general, [nice] formula that gives the exact result. (Lancelot Link gave it to me when I was studying AOPS.)
\(\sum \limits_{k=0}^{m}\binom{n}{k} (-1)^k \binom{p – s*k – 1}{p – s*k – n} \leftarrow \text {where p is the point (sum) target, n is the # of die, }\\ \hspace {47mm} \text{s is the number of sides, and } m=\lfloor \dfrac{p-n}{s}\rfloor \\\)
This formula is from J. V. Uspensky’s Introduction to Mathematical Probability (1937). There is an extensive derivation here.
Here’s the Wolfram code for this fromula:
This is compliments of Lancelot Link. (I think that Chimp could program Wolfram to bring back a burning sample of brimstone from heII.)
The variables are preset. Paste this into Wolfram Alpha.
p=20,n=10,s=6; sum (binom(n,k)*(-1)^k*(binom((p-s*k-1),( p-s*k-n))) from k=0 to Floor[(p - n)/s])
Or click here.
Now, take your [much, much] arrogant and useless commentary and shove it up your [sad, sad] áss!
+2446 I understand, I think! I will simplify \(36\div12+5-(4)(6)+(16-8)\).
| \(36\div12+5-(4)(6)+(16-8)\) | First, do what is inside of the parentheses. |
| \(36\div12+5-4*6+8\) | Let's simplify both the division and multiplication now. |
| \(3+5-24+8\) | Now, evaluate from left to right. |
| \(8-24+8\) | |
| \(-16+8\) | |
| \(-8\) | |
It is possible that, after all of this solving, that you have a question. If you do, just ask! I (or any other active member of this forum) will be happy to answer. Don't hesitate.
+2446 Thanks for responding quickly! I will evaluate it, to your choice, as the latter one of \(f(7)=\frac{6(x+3)}{11(x+7)}\).
| \(f(7)=\frac{6(x+3)}{11(x+7)}\) | First, replace every instance of x with 7, as x=7 by the given information. |
| \(\frac{6(7+3)}{11(7+7)}\) | Simplify within the parentheses first. |
| \(\frac{6*10}{11*14}\) | Now, simplify both the numerator and the denominator. However, to make it easier computationally, notice how 14 and 10 have a common factor of 2, so we can factor it out, so we don't have to do 11*14, which some have not memorized yet. |
| \(\frac{6*5}{11*7}\) | Now, simplify the numerator and denominator. |
| \(\frac{30}{77}\) | |
+2446 #4
I will evaluate \(6^2+4(17+5)-50-(36-10)\) now
| \(6^2+4(17+5)-50-(36-10)\) | First, evaluate inside of the parentheses first to adhere to the order of operations. |
| \(6^2+4*22-50-26\) | Do any exponents next, in this case 6^2. |
| \(36+4*22-50-26\) | Now, do the multiplication since that is prioritized above addition and subtraction. |
| \(36+88-50-26\) | Now, compute from left to right. |
| \(124-50-26\) | |
| \(74-26\) | |
| \(48\) | |
+2446 Time to do #2. Now, I will simplify the expression, as I interpret it to be, \(\frac{6*4}{7}+\left(\frac{5}{6}+\frac{1}{3}\right)\):
Of course, the parentheses are unnecessary in this expression as any order of adding is allowed by the associative property of addition. There is also one for multiplication.
| \(\frac{6*4}{7}+\frac{5}{6}+\frac{1}{3}\) | Simplify the numerator of the leftmost fraction. |
| \(\frac{24}{7}+\frac{5}{6}+\frac{1}{3}\) | The LCD in this case is 42, so convert all fractions to have this common denominator. |
| \(\frac{24}{7}*\frac{6}{6}=\frac{20*6+4*6}{42}=\frac{144}{42}\) | |
| \(\frac{5}{6}*\frac{7}{7}=\frac{35}{42}\) | |
| \(\frac{1}{3}*\frac{14}{14}=\frac{14}{42}\) | Now, add the fractions together. |
| \(\frac{144}{42}+\frac{35}{42}+\frac{14}{42}=\frac{193}{42}\) | |
+2446 I'll solve each one in a different comment.
#1
Simplify the expression \(\frac{5}{9}+\frac{2}{3}+\frac{1}{4}+5\).
| \(\frac{5}{9}+\frac{2}{3}+\frac{1}{4}+5\) | Get every fraction in a form such that all the fractions have a common denominator. The LCD of 9,3, and 4 is 36. Therefore, out goal is to convert every fraction that it has a denominator of 36. |
| \(\frac{5}{9}*\frac{4}{4}=\frac{20}{36}\) | |
| \(\frac{2}{3}*\frac{12}{12}=\frac{24}{36}\) | |
| \(\frac{1}{4}*\frac{9}{9}=\frac{9}{36}\) | Reinsert all of these fractions into the orginal expression. |
| \(\frac{20}{36}+\frac{24}{36}+\frac{9}{36}+5\) | Add the fractions together. |
| \(5\frac{53}{36}\) | Now, convert this mixed number into an improper fraction. |
| \(5\frac{53}{36}=\frac{36*5+53}{36}=\frac{233}{36}\) | |
+2446 One of the greatest ways to detect an error in solving is to do it yourself. I, too, have trouble spotting errors that others make--even if it is very obvious. However, if you do the problem and compare your answer with the given result, it is much easier to spot the error. I will also solve \(\frac{3}{x}+\frac{6}{x+3}=\frac{8}{x+1}\) because it is good practice.
| \(\frac{3}{x}+\frac{6}{x+3}=\frac{8}{x+1}\) | First, the solver in this problem states that the LCD is \(x(x+3)(x+1)\), which is correct. Now, let's multiply both sides by the LCD | ||
| \(\frac{3x(x+3)(x+1)}{x}+\frac{6x(x+3)(x+1)}{x+3}=\frac{8x(x+3)(x+1)}{x+1}\) | Simplify every fraction. Notice that, in every fraction, you can simplify it such that the fraction disappears. Let's do that. | ||
| \(3(x+3)(x+1)+6x(x+1)=8x(x+3)\) | Right away, I already see a disagreement between my answer and the solver's answer. They are both different. This is the error. Since I like going the extra mile of solving, I will do that! First, I'll distribute 3(x+3) first. | ||
| \((3x+9)(x+1)+6x(x+1)=8x(x+3)\) | You'll notice that it is not worth looking at the method that the solver uses, as I have already spotted the error. I can now use my own method to solve. Remember that \(ac+bc=(a+b)c\) by the inverse of the distributive property. I will use this in the next step. | ||
| \((6x+3x+9)(x+1)=8x(x+3)\) | Combine like terms in the first set of parentheses and subtract \(8x(x+3)\) from both sides of the equation. | ||
| \((9x+9)(x+1)=8x(x+3)\) | Unfortunately, I do not see a way to use the above reverse-distributive-property trick to simplify any further. I can factor out a 9 from both sides. | ||
| \(9(x+1)(x+1)=8x(x+3)\) | Now, do \((a+b)^2=a^2+2ab+b^2\). | ||
| \(9(x^2+2x+1)=8x(x+3)\) | Now, expand both sides. | ||
| \(9x^2+18x+9=8x^2+24x\) | Now, subtract 8x^2+24 from both sides. | ||
| \(x^2-6x+9=0\) | This trinomial happens to be a perfect-square trinomial. Let's use that to our advantage. | ||
| \((x-3)^2=0\) | Take the square root of both sides. Remember that taking the square root means taking the absolute value. | ||
| \(|x-3|=0\) | The absolute value splits your answer into the negative and positive answers. Luckily, the absolute value of 0 is always 0--whether positive or negative. | ||
| For the left equation, add 3. For the right equation, divide by -1. | ||
| For the second equation, we can stop because we already calculated the value for x when x-3=0. Of course, it's 3. | ||
This value for x is outside of the undefined singularity point, so this answer is correct.
For your information, you were solving this equation using the quadratic formula incorrectly. Your first line using the quadratic equation is \(x = {-6 \pm \sqrt{(-6)^2-4(1)(4)} \over 2(1)}\), but it should be \(x = {\textcolor{red}{-}(-6) \pm \sqrt{b^2-4ac} \over 2a}\). Also, \(-3\) happens to be an undefined singularity point because it is a solution to \(x+3=0\).
Another thing to mention is that I agree with Cphill that it is easier to simplify the left-hand side first, but it might be harder to detect the error if you do not do the problem as the solver did it.
