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How do you solve this using the order of operations. Please and thank you!27/9*3+6

How do you solve this using order of operations?

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Order of Operations Calculator - Math Is Fun

Order of Operations Calculator. Type in your sum to see how to solve it step by step. Examples: 2+3*4 or 3/4*3 . ... You would do the calculations from the top down ...
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Order of Operations - Free Math Help

The order of operations will allow you to solve this problem the right way. The order is this: ... using the order of operations at all times. $$ 5+(6*2) ...
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The Order of Operations: PEMDAS | Purplemath

Explains the order of operations ... "PEMDAS", which is turned into the mnemonic phrase "Please ... The order of operations was settled upon in order to ...
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What operations would you use to solve 3x-4=26. Please write ...

What operations would you use to solve 3x-4=26. Please write the operations in the order you would use them. - 4345944. 1. Log in Join now Katie; a few seconds ago;
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Order of Operations | Wyzant Resources

The Order of Operations is a standard that defines the order in which you ... This standard is critical to simplifying and solving ... use the Order of Operations ...
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6 ÷ 2 (2 + 1) = CORRECT WAY TO SOLVE - YouTube

Apr 30, 2011 · This is the correct way to solve the equation. ... Order of Operations 48/2[9+3] - Duration: ... 6:27. tecmath 5,126,047 views.

Algebraic Expressions, Order of Operations/P.E.M.D.A.S. I

Evaluating Algebraic Expressions Order of Operations/P.E.M.D.A.S. first of ... Evaluating Algebraic Expressions Order of Operations/P.E ... Please Excuse My Dear ...
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What is PEMDAS? - Definition, Rule & Examples - Video ...

This order may also be memorized using the phrase Please ... helps you remember the order of operations ; Use ... is PEMDAS? - Definition, Rule & Examples ...
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How to Simplify Math Expressions: 13 Steps (with Pictures)

How to Simplify Math Expressions. ... A handy acronym you can use to remember this is "Please excuse ... After proceeding through the order of operations, you should ...
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Suggested Questions And Answer :

How do you solve this using the order of operations. Please and thank you!27/9*3+6

quesshun: 27/9*3 +6 =(27/9)*3+6= 3*3+6 =9+6=15
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solve for r in this equation 2000[1+r(51/365)]=2038.11

alright, to isolate r we'll have to use inverse operations  here is your original equation to refer back to: 2000[1 + r(51/365)] = 2038.11 first, let's divide both sides by 2000, so we'll have 1.02 (roughly - i rounded up) now the equation is 1 + r(51/365) = 1.02 subtract 1 from both sides now we have approximately .02 51/365 is roughly equal to .14 with our equation set up as r(51/365) = .02, we'll divide both sides by 51/365 now we have, as a very long decimal, .1363740196, which we can round to .14 or, if you want, .136 r = .136 now we'll go back and check the work 2000[1 + .136(51/365)] = 2038.11 multiply .136 with 51/365 first, in accordance with the order of operations, to get .019 now add 1 to get 1.019 now multiply this by 2000 the answer comes out to 2038.11 so r is approximately equal to 0.136
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Solve, using the 5-step plan.

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How to solve (3x+22)=(6x-8)?

How to solve (3x+22)=(6x-8)? Our lesson is now about Parallel Lines and we have to solve first before using the theorems. And I don't know how to solve it. Please help me. Oh, and the teacher said thar the working equations is m<1 = m<2. Thank you so much 3x + 22 = 6x - 8 3x + 22 - 3x = 6x - 8 - 3x 22 = 3x - 8 22 + 8 = 3x - 8 + 8 30 = 3x 30 / 3 = 3x / 3 10 = x
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How to solve for: 3[4-3(6-7)]

Question: How to solve for: 3[4-3(6-7)] There is a thing called operator precedence. This tells you in what order algebraic operators should be applied. The order is, 1) exponents and roots 2) multiplication and division 3) addition and subtraction In addition to that, where brackets are used, these are evaluated first. So although a set of brackets is not an operator, they take highest priority. Innermost brackets should be evaluated first. Our expression is: 3[4-3(6-7)]   -- evaluate the innermost bracket first. 3[4 - 3*(-1)]  -- multiplication takes precedence, so do the 3*(-1) first. 3[4 + 3]    ----- finish evaluating the bracket with the addition operation. 3*7     ---------- complete the expression's evaluation with a final addition. 21 Answer: 21
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12divided by 6 times 3 minus 5

Problem: 12divided by 6 times 3 minus 5 Please show us the order of operations and the answer.  thank you Because there are no parentheses to force the order of operation, take everything in the order they are presented, understanding that all multiplication and division operations are done before any addition and subtraction. 12 / 6 * 3 - 5 First: 12 / 6 = 2 2 * 3 - 5 Second: 2 * 3 = 6 Third: 6 - 5 Answer: 1 If there had been parentheses around the 6 * 3, the answer would have been different. You would have had 12 / 18, giving 2/3 - 5. Also, if there had been parentheses around the 3 - 5, the answer would have been different. In that case, you would have had 2 * -2.  
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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3X8 -3+ 9 =

3X8 -3+ 9 = Using the order of operations and working left to right, how do you solve this problem?   3 x 8 - 3 + 9 = 24 - 3 + 9 = 21+9 = 30  
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12 ÷ 2 + 4 – 2 × 3 = ?

The answer is four. Just remember Please Excuse My Dear Aunt Sally = Parenthesis Exponents Multiply Divide Add Subtract. That is the easiest way to remember that.
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how can you get the answer 7 while only using the number 2,0,1, and 2?

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