SIGMA 12^4 (.56-.09 SQ.ROOT OF 11.8)(.1+.9*.5)\n\nWhat do I do with the SIGMA?\nAnd I multiply that square root, right (There is no multiplication sign, just a space)?
216390
As best as I can translate, the equation you are intending looks like this:\n\nΣ [12^4 (.56-.09 √¯(11.8)) (.1+.9*.5)]\n\nIn which case, the problem is incomplete. The summation function, Σ, requires that there be a variable, such as x, in the equation. It also requires that the variable x have set parameters and that these parameters are shown above and below the summation symbol. \n\nFor example:\n\nx=5\n Σ [12^4 (.56-.09x √¯(11.8)) (.1+.9*.5)]\nx=2\n\nYou ignore Σ for the moment, and find the solution of the rest of the expression for each of the values of x indicated (2, 3, 4, & 5). You will then have just four numbers; in other words, all those messy operations will be gone. Σ means you add those numbers now.\n\nIn all cases, the order of operations are as follows:\n\nParentheses, Exponents (√¯ is just an exponent of ½), Multiplication/Division, Addition/Subtraction, Summation\n\nAddition is not Summation and shouldn't be mistaken for it.\n\n\nTo answer your other question, when there is no other operation between numbers, or parentheses, the default operation is multiplication. Therefore, after finding the square root of 11.8, you do indeed multiply that answer with .09.\n\n\nP.S. All the mathematical symbols in this explanation were found using my computer's character map. You can do a search on your computer to find yours. Click advanced. The bottom field allows for searches. (My version of the square root symbol is the square root symbol followed by a macron.)
559020
u r probably working on order of operations, this might help:\nPlease (parantheses)\nExcuse (exponents)\nMy  (multiplication)\nDear (division)\nAunt (addition)\nSally (subtraction)\n\n1. 7000 x 400000 = 2800000000\n2.  1/144 = x^2 so x = +/- 1/12\n3.  -3 +/- the square root of 2\n4.  5t^4 - 7t^3 + 6t -2\n5.  4x^4 - 16x^2 +16
504431
what it means is this 4 plus negative .01 and with the other it's 5 plus negative .04 its simple. if you know what it means and if you took notes look at them. it may help a little
392553
First you find the square root of both the numerator and the denominator which would end up being √5/√8. This would be a fine answer, but in my math classes teachers have had rules against keeping a radical in the denominator. so, to get rid of that multiply √5/√8 by √8/√8, which gives you √40/√64, or √40/8.
391904
Alright. i=1-5,because the number on the bottom is the first number you will put in the expresion, and the number on top is the last number.and you put all the number inbetween in the expression too. So i will equal 1, 2, 3, 4, 5. Plug each one into the expression next to E. Now you have 5 answers. Sigma means something like "add" so you add all 5 answers together, and that is the final answer.
49813
No idea here too... what i know is that it could be simplified to 34x + 40. Dunno about the blank (____)
471319
c to f:\nn*1.8; add 32\n\nf to c:\nn-32; multiply by 0.5555
551422
We use the symbol above the number 6 on the keyboard to show exponents.  For example, 2 squared is typed 2^2 = 4.\n\nWhy don't you use the symbol ^ and retype your question correctly?\n\nGuido
491436
S.D. = ((Sigma xi^2 - (Sigma xi / n)) / (n-1))^1/2
482691
Ever hear of a spacebar? And you don't write "square" to indicate squaring, you write ² or ^2.\n\nAnyway, onto the problem:\n\nx² - 2xy + y² - 5x + 7y ≡ 0 mod 17\nx² - (2y+5)x + y² + 7y ≡ 0 mod 17\nx² - (2y+5)x ≡ -y² - 7y mod 17\nNow we complete the square. First we find (2y+5)/2 mod 17, which is y+11 (since 11*2 ≡ 5 mod 17). Now, we square it, to yield y²+5y+2 (again, mod 17). \nx² - (2y+5)x + y² + 5y + 2 ≡ -y² - 7y + y² + 5y + 2 mod 17\n(x-y+11)² ≡ -2y + 2 mod 17\nx ≡ y - 11 ± √(-2y+2) mod 17\n\nNow, we substitute this into the other congruence:\n\nx² - 3xy + 2y² + x - y ≡ 0 mod 17\n(y - 11 ± √(-2y+2))² - 3(y - 11 ± √(-2y+2))y + 2y² + y - 11 ± √(-2y+2) - y ≡ 0 mod 17\n\nMultiplying everything out:\n\ny² - 7y ± 2y√(-2y+2) + 4 ± 12√(-2y+2) - 3y² + 33y ± 14y√(-2y+2) + 2y² + y - 11 ± √(-2y+2) - y ≡ 0 mod 17\n\nAnd simplifying:\n\n9y - 7 ± (16y+13)√(-2y+2) ≡ 0 mod 17\n± (16y+13)√(-2y+2) ≡ 7-9y mod 17\n\nSquaring both sides:\n\n(y² + 8y + 16)(-2y+2) ≡ 13y² - 7y + 15 mod 17\n-2y³ - 16y² - 15y + 2y² + 16y + 15 ≡ 13y² - 7y + 15 mod 17\n-2y³ - 14y² + y + 15 ≡ 13y² - 7y + 15 mod 17\n-2y³ - 10y² + 8y ≡ 0 mod 17\n-y(2y² + 10y - 8) ≡ 0 mod 17\nSince 17 has no zero divisors, this implies that y≡0 mod 17 or  2y² + 10y - 8 ≡ 0 mod 17. In the latter case:\n\n2y² + 10y - 8 ≡ 0 mod 17\ny² + 5y - 4 ≡ 0 mod 17\ny²+5y+2 ≡ 6 mod 17\n(y+11)² ≡ 6 mod 17\n\nHowever, 6 has no square roots mod 17, so this is impossible. So y≡0 mod 17. We may now go back and solve for x:\n\nx ≡ y - 11 ± √(-2y+2) mod 17\nx ≡ - 11 ± √2 mod 17\nx ≡ -11 ± 6 mod 17\nx ≡ 6 ± 6 mod 17\nx ≡ 0 mod 17 or x≡12 mod 17\n\nWe have two possible solutions - it is best that we test them on your original congruences, which are:\n\nx² - 2xy + y² - 5x + 7y ≡ 0\nx² - 3xy + 2y² + x - y ≡ 0\n\nClearly, these are both satisfied by the solution x≡0, y≡0. However, we have a second possible solution, which is x≡12, y≡0. Testing it on the first congruence gives us:\n\nx² - 2xy + y² - 5x + 7y ≡ 0\n8 - 9 ≡ 0 mod 17\n\nWhich is clearly not true. So the only way the given congruence relations can hold is if x≡0 and y≡0. Substituting this solution into xy-12x+15y trivially yields that this expression is congruent to zero, and we are done.
383354
You didn't answer this correctly. if the number inside the square root (-320) is negative, then it's an imaginary number.\n\nYou also shouldn't have a plus or minus sign to the left of the square root at this point of the equation if you've done it correctly.\n\nThere are two answers to this question once you've done it correctly, unless the question asks you to measure something, in which case, the only correct answer is the positive value. Try posting the actual question you were given and what you came up with. Perhaps I can help you there.\n\nGood luck
439306
goto: http://www.google.com\n\nUse the search queery: "What is the square root of"  and then insert your numbers.
539135
That just means that keep it in the square root instead of simplifying it.\nEx: Keep it as square root of 2 instead of 1.414\n\nSo if you're finding the square root of 27, for example.  You can split up 27 into 9*3.  Square root of 9 is 3, but 3 is not a perfect square.  So your answer would be : 3 square root of 3.  Instead of 5.196
6587
let me give you the distance formula: the square root of ( [X2 - X1]squared + [Y2 - Y1]squared )\n\nwhat are X2 and Y1 and stuff like that?  well, your coordinates are (-7,6) and (-1,1).  You assign one set of coordinates (it doesn't matter wich) to be x1 and y1.  the other set will be x2 and y2.  So, for example, in the point (-7,6) your -7 would be x1 and your 6 would be y1.  then your -1 would be x2 and your - would be y2.  make sense?  so once you have assigned your x1, x2, y1, and y2, then you just plug them into the distance formula:\n\nthe square root of ( [-1 - -7]squared + [1-6]squared ).\n\nyou solve that, and there's your answer!  I would use a scientific calculator to do the squaring and square rooting, and I assume you know which buttons do that.
391845
If this is a problem in a math class, the answer provided by abcde12345 is correct. He shows you the technique to simplify it, which is to multiply numerator and denominator by the "conjugate" of the denominator (which just means you change the sign of the square root part, so you get (3 - sqrt(5)).\n\nIf this is a problem in physics or some other class, and all you need is a value, then follow the suggestion of some others who recommend using a calculator.
612505
[ (50000 x 2.4)+(25000 x .8) ] / (50000+250000)=1.8\n\nThe remainder of the info in the problem is not necessary.
370130
You figure the square root of each number separately...you've probably learned that in class alredy (how to do that)...then you take reach result, and multiply them together for your final answer.\n\nUse your textbook for reference...good luck!