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You have to get rid of the parenthesis before moving on
It is gotten rid of, as soon as the addition is done.
It is gotten rid of, as soon as the addition is done.
It is gotten rid of, as soon as the addition is done.
The parenthesis is there, so that you dont multiply 2 with 9
48 : 2 * 9 + 3 => then you would add 3 last
ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
It doesnt....its gone...I dont know why you would assume that it would remain.
There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
Which is also explained in the earlier post. XY = X(Y) = X*Y
............i feel so repetitive right now
You have to get rid of the parenthesis before moving on
It is gotten rid of, as soon as the addition is done.
It is gotten rid of, as soon as the addition is done.
It is gotten rid of, as soon as the addition is done.
The parenthesis is there, so that you dont multiply 2 with 9
48 : 2 * 9 + 3 => then you would add 3 last
ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
It doesnt....its gone...I dont know why you would assume that it would remain.
There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
Which is also explained in the earlier post. XY = X(Y) = X*Y
............i feel so repetitive right now
.... By parentheses it means YOU DO WHAT IS INSIDE THE PARENTHESES FIRST.Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
.... By parentheses it means YOU DO WHAT IS INSIDE THE PARENTHESES FIRST.Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
There is actually an implied/mostly invisible parenthesis around any terms thats why they are "terms". so the problem shoud turn into (4/(2)*(12)Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
There is actually an implied/mostly invisible parenthesis around any terms thats why they are "terms". so the problem shoud turn into (4/(2)*(12)Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
BUT THE OG PROBLEM IS 48÷2(9+3)Originally Posted by VietStar
Parentheses first.
48/2(9+3).
48/2(12).
Then divide. If the equation has both a division and multiplication sign, such as this one, then you do whatever occurs first, first. In this problem, division occurs first, so you divide. If it was instead a multiplication sign first, then you would multiply before dividing.
You get:
24(12)
Then multiply it.
Final Answer:
288
This is such an easy equation, but I can understand why and how someone would mess it up. It's just funny how the people who have the wrong answers cannot face the fact that they're wrong, and yet, call others stupid and dumb. They are too ignorant to realize their mistakes.
Sorry for the big font, but after 117 pages of nonsense and arguments, I just wanted to get the point across and show people how do this equation correctly in an easy/understandable format.
BUT THE OG PROBLEM IS 48÷2(9+3)Originally Posted by VietStar
Parentheses first.
48/2(9+3).
48/2(12).
Then divide. If the equation has both a division and multiplication sign, such as this one, then you do whatever occurs first, first. In this problem, division occurs first, so you divide. If it was instead a multiplication sign first, then you would multiply before dividing.
You get:
24(12)
Then multiply it.
Final Answer:
288
This is such an easy equation, but I can understand why and how someone would mess it up. It's just funny how the people who have the wrong answers cannot face the fact that they're wrong, and yet, call others stupid and dumb. They are too ignorant to realize their mistakes.
Sorry for the big font, but after 117 pages of nonsense and arguments, I just wanted to get the point across and show people how do this equation correctly in an easy/understandable format.
Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
No it shouldn't because AGAIN, w[color= rgb(255, 255, 255)]hat you fail and refuse to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved firs[/color][color= rgb(255, 255, 255)]t before you proceed with the rest of the problem. [/color]Originally Posted by usainboltisfast
There is actually an implied/mostly invisible parenthesis around any terms thats why they are "terms". so the problem shoud turn into (4/(2)*(12)Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
No it shouldn't because AGAIN, w[color= rgb(255, 255, 255)]hat you fail and refuse to understand is that 2 is the common factor of 18 and 6 (also 3), hence 2(9+3). If you choose 3 as the common factor it will be like this 3(6+2). Therefore the number next to the parenthesis must be resolved firs[/color][color= rgb(255, 255, 255)]t before you proceed with the rest of the problem. [/color]Originally Posted by usainboltisfast
There is actually an implied/mostly invisible parenthesis around any terms thats why they are "terms". so the problem shoud turn into (4/(2)*(12)Originally Posted by balloonoboy
After you add what's in the parenthesis, there is still a parenthesis though.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
You have to get rid of the parenthesis before moving on.
So, what you gonna do? 48/2x12? That is not the same as 48/2(12).
All you dudes asking your professors, ask them why the parenthesis remains when the whole point of P is to take care of the parenthesis.
Now if you had a problem that read 48/2x(9+3)...it would be easy to get rid of the parenthetical. Just add what's inside. There is no coefficient (greater than 1) attached to it. The problem at hand, there is.
Originally Posted by kingcrux31
BUT THE OG PROBLEM IS 48÷2(9+3)Originally Posted by VietStar
Parentheses first.
48/2(9+3).
48/2(12).
Then divide. If the equation has both a division and multiplication sign, such as this one, then you do whatever occurs first, first. In this problem, division occurs first, so you divide. If it was instead a multiplication sign first, then you would multiply before dividing.
You get:
24(12)
Then multiply it.
Final Answer:
288
This is such an easy equation, but I can understand why and how someone would mess it up. It's just funny how the people who have the wrong answers cannot face the fact that they're wrong, and yet, call others stupid and dumb. They are too ignorant to realize their mistakes.
Sorry for the big font, but after 117 pages of nonsense and arguments, I just wanted to get the point across and show people how do this equation correctly in an easy/understandable format.
[font=Arial, Helvetica, sans-serif]
[/font]
Originally Posted by kingcrux31
BUT THE OG PROBLEM IS 48÷2(9+3)Originally Posted by VietStar
Parentheses first.
48/2(9+3).
48/2(12).
Then divide. If the equation has both a division and multiplication sign, such as this one, then you do whatever occurs first, first. In this problem, division occurs first, so you divide. If it was instead a multiplication sign first, then you would multiply before dividing.
You get:
24(12)
Then multiply it.
Final Answer:
288
This is such an easy equation, but I can understand why and how someone would mess it up. It's just funny how the people who have the wrong answers cannot face the fact that they're wrong, and yet, call others stupid and dumb. They are too ignorant to realize their mistakes.
Sorry for the big font, but after 117 pages of nonsense and arguments, I just wanted to get the point across and show people how do this equation correctly in an easy/understandable format.
[font=Arial, Helvetica, sans-serif]
[/font]
This next example displays an issue that almost never arises but, when it does, there seems to be no end to the arguing.
Simplify 16 ÷ 2[8 â 3(4 â 2)] + 1.
16 ÷ 2[8 â 3(4 â 2)] + 1
= 16 ÷ 2[8 â 3(2)] + 1
= 16 ÷ 2[8 â 6] + 1
= 16 ÷ 2[2] + 1 (**)
= 16 ÷ 4 + 1
= 4 + 1
= 5
The confusing part in the above calculation is how "16 divided by 2[2] + 1" (in the line marked with the double-star) becomes "16 divided by 4 + 1", instead of "8 times by 2 + 1". That's because, even though multiplication and division are at the same level (so the left-to-right rule should apply), parentheses outrank division, so the first 2 goes with the [2], rather than with the "16 divided by". That is, multiplication that is indicated by placement against parentheses (or brackets, etc) is "stronger" than "regular" multiplication.
It was posted before and dudes still arguing. Wrap it up.
Spoiler [+]Elizabeth Stapel's resume is loooong
Elizabeth Stapel's ResumÃ[emoji]169[/emoji]
Teaching Experience
Western International University, Phoenix, Arizona
October 1997 through June 2006: teaching prealgebra, algebra, and business math to students of diverse backgrounds in a business environment.
Winter 2003 through Spring 2004: helped design an introductory proseminarthat providesenteringstudents with information and encouragement regarding college success.
12 June 1999: helped lead a faculty study group on the subject of effective lecturing.
Chandler-Gilbert Community College, Chandler, Arizona
Spring and Fall 1999, Spring 2000 semesters: teaching algebra
South Mountain Community College, Phoenix, Arizona
Spring and Summer 1996, Fall 1997, Fall 1998, and Spring 1999 semesters: teaching high school students in a collaborative pilot algebra program; teaching elementary algebra classes; teaching/piloting a reform calculus class.
Arizona State University, Tempe, Arizona
Spring 1996 semester: teaching elementary algebra and elementary linear algebra; assisting department's book-search committee in choosing a new algebra text.
Simply Mathematics Tutoring, Ballwin, Missouri
February 1995 to June 1995: tutoring middle school, high school, and college students in various areas of mathematics.
Saint Louis University, Saint Louis, Missouri
Fall 1994 semester: teaching an evening calculus course for non-traditional, multi-cultural students.
St. Louis Community College of Florissant Valley, Saint Louis, Missouri
Fall 1994 semester: teaching non-traditional, multi-cultural students in remedial arithmetic and beginning algebra.
Southwestern Bell Telephone and Florissant Valley Community College, Saint Louis, Missouri
August 1994: teaching a weekend algebra review session for Southwestern Bell employees, preparing them for a college-administered placement exam (a sufficient score on the exam entered students in a special study program, leading to a promotion within the company).
Sylvan Learning Center, Creve Coeur, Missouri
April 1994 to July 1994: tutoring students of all school grades and of all abilities in Reading, Mathematics, Study Skills, and ACT Prep Programs.
Washington University, Saint Louis, Missouri
August 1992 to May 1994: leading weekly supplementary calculus help sessions, grading quizzes; collecting and assessing student feedback and opinion.
Fall 1993 semester: leading supplementary calculus help-sessions in student dorms.
Wright State University, Dayton, Ohio
August 1991 to June 1992: teaching precalculus algebra students; staffing Math Help Room.
August 1990 to June 1991: proctoring (group tutoring and teaching) for an innovative self-study remedial math program.
March 1990 to August 1991: grading homework for math department; proctoring tests; assisting in developing a departmental homework grading policy.
Summer 1990 quarter: staffing Math Help Room: assisting students in all classes through calculus.
January 1989 to March 1990: working for the University Tutoring Division: tutoring all lower-level math classes and all classifications of students, including disabled and LD.
This next example displays an issue that almost never arises but, when it does, there seems to be no end to the arguing.
Simplify 16 ÷ 2[8 â 3(4 â 2)] + 1.
16 ÷ 2[8 â 3(4 â 2)] + 1
= 16 ÷ 2[8 â 3(2)] + 1
= 16 ÷ 2[8 â 6] + 1
= 16 ÷ 2[2] + 1 (**)
= 16 ÷ 4 + 1
= 4 + 1
= 5
The confusing part in the above calculation is how "16 divided by 2[2] + 1" (in the line marked with the double-star) becomes "16 divided by 4 + 1", instead of "8 times by 2 + 1". That's because, even though multiplication and division are at the same level (so the left-to-right rule should apply), parentheses outrank division, so the first 2 goes with the [2], rather than with the "16 divided by". That is, multiplication that is indicated by placement against parentheses (or brackets, etc) is "stronger" than "regular" multiplication.
It was posted before and dudes still arguing. Wrap it up.
Spoiler [+]Elizabeth Stapel's resume is loooong
Elizabeth Stapel's ResumÃ[emoji]169[/emoji]
Teaching Experience
Western International University, Phoenix, Arizona
October 1997 through June 2006: teaching prealgebra, algebra, and business math to students of diverse backgrounds in a business environment.
Winter 2003 through Spring 2004: helped design an introductory proseminarthat providesenteringstudents with information and encouragement regarding college success.
12 June 1999: helped lead a faculty study group on the subject of effective lecturing.
Chandler-Gilbert Community College, Chandler, Arizona
Spring and Fall 1999, Spring 2000 semesters: teaching algebra
South Mountain Community College, Phoenix, Arizona
Spring and Summer 1996, Fall 1997, Fall 1998, and Spring 1999 semesters: teaching high school students in a collaborative pilot algebra program; teaching elementary algebra classes; teaching/piloting a reform calculus class.
Arizona State University, Tempe, Arizona
Spring 1996 semester: teaching elementary algebra and elementary linear algebra; assisting department's book-search committee in choosing a new algebra text.
Simply Mathematics Tutoring, Ballwin, Missouri
February 1995 to June 1995: tutoring middle school, high school, and college students in various areas of mathematics.
Saint Louis University, Saint Louis, Missouri
Fall 1994 semester: teaching an evening calculus course for non-traditional, multi-cultural students.
St. Louis Community College of Florissant Valley, Saint Louis, Missouri
Fall 1994 semester: teaching non-traditional, multi-cultural students in remedial arithmetic and beginning algebra.
Southwestern Bell Telephone and Florissant Valley Community College, Saint Louis, Missouri
August 1994: teaching a weekend algebra review session for Southwestern Bell employees, preparing them for a college-administered placement exam (a sufficient score on the exam entered students in a special study program, leading to a promotion within the company).
Sylvan Learning Center, Creve Coeur, Missouri
April 1994 to July 1994: tutoring students of all school grades and of all abilities in Reading, Mathematics, Study Skills, and ACT Prep Programs.
Washington University, Saint Louis, Missouri
August 1992 to May 1994: leading weekly supplementary calculus help sessions, grading quizzes; collecting and assessing student feedback and opinion.
Fall 1993 semester: leading supplementary calculus help-sessions in student dorms.
Wright State University, Dayton, Ohio
August 1991 to June 1992: teaching precalculus algebra students; staffing Math Help Room.
August 1990 to June 1991: proctoring (group tutoring and teaching) for an innovative self-study remedial math program.
March 1990 to August 1991: grading homework for math department; proctoring tests; assisting in developing a departmental homework grading policy.
Summer 1990 quarter: staffing Math Help Room: assisting students in all classes through calculus.
January 1989 to March 1990: working for the University Tutoring Division: tutoring all lower-level math classes and all classifications of students, including disabled and LD.
Originally Posted by balloonoboy
Okay, this will end the thread. The answer is two, not 288 or 8.6 (much to my chagrin).
http://www.purplemath.com/modules/orderops2.htm
This next example displays an issue that almost never arises but, when it does, there seems to be no end to the arguing.
Simplify 16 ÷ 2[8 â 3(4 â 2)] + 1.
16 ÷ 2[8 â 3(4 â 2)] + 1
= 16 ÷ 2[8 â 3(2)] + 1
= 16 ÷ 2[8 â 6] + 1
= 16 ÷ 2[2] + 1 (**)
= 16 ÷ 4 + 1
= 4 + 1
= 5
The confusing part in the above calculation is how "16 divided by 2[2] + 1" (in the line marked with the double-star) becomes "16 divided by 4 + 1", instead of "8 times by 2 + 1". That's because, even though multiplication and division are at the same level (so the left-to-right rule should apply), parentheses outrank division, so the first 2 goes with the [2], rather than with the "16 divided by". That is, multiplication that is indicated by placement against parentheses (or brackets, etc) is "stronger" than "regular" multiplication.
Originally Posted by balloonoboy
Okay, this will end the thread. The answer is two, not 288 or 8.6 (much to my chagrin).
http://www.purplemath.com/modules/orderops2.htm
This next example displays an issue that almost never arises but, when it does, there seems to be no end to the arguing.
Simplify 16 ÷ 2[8 â 3(4 â 2)] + 1.
16 ÷ 2[8 â 3(4 â 2)] + 1
= 16 ÷ 2[8 â 3(2)] + 1
= 16 ÷ 2[8 â 6] + 1
= 16 ÷ 2[2] + 1 (**)
= 16 ÷ 4 + 1
= 4 + 1
= 5
The confusing part in the above calculation is how "16 divided by 2[2] + 1" (in the line marked with the double-star) becomes "16 divided by 4 + 1", instead of "8 times by 2 + 1". That's because, even though multiplication and division are at the same level (so the left-to-right rule should apply), parentheses outrank division, so the first 2 goes with the [2], rather than with the "16 divided by". That is, multiplication that is indicated by placement against parentheses (or brackets, etc) is "stronger" than "regular" multiplication.
Originally Posted by kingcrux31
BUT THE OG PROBLEM IS 48÷2(9+3)Originally Posted by VietStar
Parentheses first.
48/2(9+3).
48/2(12).
Then divide. If the equation has both a division and multiplication sign, such as this one, then you do whatever occurs first, first. In this problem, division occurs first, so you divide. If it was instead a multiplication sign first, then you would multiply before dividing.
You get:
24(12)
Then multiply it.
Final Answer:
288
This is such an easy equation, but I can understand why and how someone would mess it up. It's just funny how the people who have the wrong answers cannot face the fact that they're wrong, and yet, call others stupid and dumb. They are too ignorant to realize their mistakes.
Sorry for the big font, but after 117 pages of nonsense and arguments, I just wanted to get the point across and show people how do this equation correctly in an easy/understandable format.
[font=Arial, Helvetica, sans-serif]
[/font]