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Could you help me please?

-7u=-12 tha,s my question?

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phrase choice - "Could you please help me" vs "Could you help ...


Could you please help me? or Could you help me please? current community. chat. English Language Learners English Language Learners Meta your communities ...
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Help Me Please! - Queendom


Can you give me some tips not to be depressed? Please help me and email me back. ... Help Me Please! Question: Hello.
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COULD YOU HELP ME, PLEASE? by LETS INSTITUTE on Prezi


COULD YOU HELP ME, PLEASE? COULD ... Could you come over here, please? = Please come here. NOW YOUR TURN MAKE A SENTENCE WITH COULD FOR EACH PICTURE, DESCRIBE THE ...
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Please could you help me out, with this 2 question, no citation


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Can You Help Me Please? I am lost: Marjorie Simmons ...


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could you help me with this or could you help me on this?


could you help me with this or could you help me on this? ... "Could you help me with this sentence, please?" So "Could you help me, please?" would be ...
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'Can you please' or 'Could you please' -- which is correct ...


"I'm sorry, 'Could you please repeat the question?'" would reflect my failure to understand the question, and your willingness to placate me by repeating the question.
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Could you help me please? - eTeacher Hebrew


Could you help me please? Transcription . hatuxal / hatuxlee la'azor lee be'vakasha? Hebrew . הֲתוּכַל לַעֲזֹר לִי ...
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Z-ro - Help me Please (lyrics) - YouTube


Dec 18, 2008 · Zro,Help me please ... This feature is not available right now. Please try again later.

Could you help me please | Chegg Tutors


Could you help me please. Asked by a Mechanical Engineering student, June 1, 2017. A Mechanical Engineering tutor answered. Nikunj G., Master in Mechanical ...
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Suggested Questions And Answer :


if x=6 and y=2 what would this equasion equall; 7+5y

No, kant help...not nuf info yu suppli 1 point...(6,2) we kan assume yu want a strate line, so y=slope*y+konst tu solv for 2 dont-nos, need more... 2 points, or suppli the slope
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simple shapes for second grade.


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Problem of the Week Cutting Challenge Grade 5 Problem 23

  Tim had 4 paper fraction pieces, each of thirds, fourths, sixths, and twelfths. He did not have enough pieces to show both 2/4 and 1/3 as equivalent fractions with the same denominator. How could he form enough pieces by making 1 cut with a scissors? My daughter had this problem and for the life of me I cannot figure it out...please help so I can understand this myself so I can explain it to my daughter. The only thing I am pretty sure about is that the same or common denominator is 12 (twelfths) but to cut paper once to show for this by producing the pieces with one cut...??? That's where I don't know what to do to demonstrate this with a single cut...Thanks in advanced for your help! 1/3 = 4/12 1/4 = 3/12 1/6 = 2/12 1/12 adding these 4/12 + 3/12 + 2/12 + 1/12 = 10/12 have to make 2/4 or 6/12 and 1/3 or 4/12 Imagine the four pieces in a circle all connected1/3 next two 1/4 next to 1/6 and 1/12 (there is 2/12 not used or missing. The Cut is between 1/3 and 1/4      
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how do you do fractions like say 4/9 in to 1/3 please tell me please!

"ELIZABETH WAS IN CHARGE OF COUNTING THE VOTES FOR CLASS PRESIDENT.CASEY RECIEVED 8/15 OF THE VOTES .HIS OPPONENT JASON RECIEVED THE REST OF THE VOTES.WHAT PARTS OF THE VOTES DID JASON RECIEVE?WHO WON THE ELECTION?" Casey received 8/15 of the votes and Jason received the rest. Since the denominator was given as 15 we will use that number to explain this. Assume that there were 15 votes total or that the votes could be seperated into 15 equal piles.  Casey received 8 of those 15 so what was left was 15-8=7.  Jason received 7/15 to Casey's 8/15. Who won?  8/15 is larger than 7/15
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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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the area of a garden is 48 m2 , what is the greatest perimater the garden could have?

Problem: the area of a garden is 48 m2 , what is the greatest perimater the garden could have? please help im stuck im in grade 5 The counter-question has to be, how unreasonable do you want to be? Here are several pairs of numbers that give the designated area: W,L    2W   2L    P          A 6,8     12 + 16 = 28    6*8 = 48 4,12     8 + 24 = 32    4*12 = 48 3,16     6 + 32 = 38    3*16 = 48 2,24     4 + 48 = 52    2*24 = 48 1,48     2 + 96 = 98    1*48 = 48 As you can see, shortening the width increases the perimeter, but makes the garden ridiculously narrow.  
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Please help

You need to see through the problem and apply whatever is necessary to reduce the number of variables till eventually there's only one to find. Remember a simple fact: if you have two variables you always need two independent equations to find them; for three variables, three equations; four variables, four equations. You use the multiplication property if it helps you to eliminate a variable between two equations. Take some examples: x+y=10, x-y=3; simply adding these two equations will eliminate y and help you find x. 2x=13 so x=6.5 and y=3.5; 2x+y=10, x-2y=10; we could double one equation or the other so as to match the coefficients of one or other of the variables; but since it's easier to add two equations rather than to subtract them, where we have a minus in one equation and a plus in the other, we would prefer to use the multiplier for the relevant variable. So we double the first equation and add to the second: 4x+2y=20 PLUS x-2y=10: 5x=30, making x=6 and y=-2. The last pair of equations could have been written: 2x+y=x-2y=10, but it's still two equations. There is no one way to solve equations, and you can save yourself a lot of stress by not assuming you have to remember a rigid technique or formula as “The Way to do it”. You'll find mathematics is more fun when you intelligently try different methods and use your natural creativity to guide you. And here's another interesting thing. Those questions about finding a missing number in a series can be tackled in many cases as solving simultaneous equations. You only need n equations to find n variables, and a series can be seen as a set of terms generated by a function y=f(x) for different values of x (the position in the series), giving different values of y (the terms in the series). f(x) is a polynomial of the type ax^n+bx^(n-1)+cx^(n-2)+... If there are four given terms n=3 and the variables are a, b, c and d; if there are 3 given terms, n=2 and the variables are a, b and c. There is always a solution, we just have to work through and find it!
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Can you please help me solve-7x(q-5)= ?

7qx-35x hope this helps
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how do i solve 1.5 times 1.8?

1.5*1.8=2.7 ............
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Please help solve 12x^3-48x^2-75x-24/6x+3

I read this as (12x^3-48x^2-75x-24)/(6x+3). Divide top and bottom by 6: 2x^3-8x^2-12.5x-4)/(x+1/2). We can use synthetic division: -1/2 | 2 -8 -12.5 -4 .........2 -1....4.5..4 .........2 -9......-8 | 0 = 2x^2-9x-8 Provided x≠-1/2 the quotient is 2x^2-9x-8 which is a simplification of the original expression.
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