Guide :

six and one sixth subtract five and twelve twenty-one

6 1/6 - 5 12/21

Research, Knowledge and Information :


How do you calculate one sixth of twelve? - Weknowtheanswer


How do you calculate one sixth of twelve? Find answers now! No. 1 Questions & Answers Place. ... one sixth subtract five and twelve ...
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Grade 4: Adding and Subtracting Fractions With Like ...


... Adding and Subtracting Fractions With Like Denominators ... When you add 6 minutes to 5 minutes, ... (one sixth) Ask: ...
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Adding and Subtracting Fractions with Pictures Pt 1 of 2 ...


Feb 03, 2014 · Adding and Subtracting Fractions with Pictures Pt 1 of 2 This series of ... but there's only one more sixth to ... Now we need to subtract one of the ...

Math Magic


zero plus fifteen one plus fourteen two plus thirteen three plus twelve one times fifteen twenty minus five forty ... a five 6 = one six ... Math Magic . Last updated ...
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Numbers in English | English Grammar Guide | EF


Numbers in English. The cardinal numbers ... twenty-one: twenty-first: 22: ... twenty-four: twenty-fourth: 25: twenty-five: twenty-fifth: 26: twenty-six: twenty-sixth ...
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Introducing the Concept - Education Place®


Introducing the Concept. ... a unit has been divided into 9 equal parts and there are five ... the student did in terms of joining one set to the other in ...
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Suggested Questions And Answer :


six and one sixth subtract five and twelve twenty-one

(6& 1/6) -(5 & 12/21) bekum (6 & 21/126) -(5 & 72/126) =(6-5)+(21-72)/126 =1 -51/126 or 75/126...0.595238095
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write a number 123,456,789,012,345,678,901,234,567,890,123 in word form

write a number 123,456,789,012,345,678,901,234,567,890,123 in word form. In word form this number would be, One hundred and twenty three nonillion, four hundred and fifty six octillion, seven hundred and eighty nine septillion, twelve sextillion, three hundred and forty five quintillion, six hundred and seventy eight quadrillion, nine hundred and one trillion, two hundred and thirty four billion, five hundred and sixty seven million, eight hundred and ninety thousand, one hundred and twenty three.
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Add three to a number, then multiply your number by 4

y=4*(x+3) ....................
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please help with describing and ordering fractions

The divisions between 0 and 1 show how a fraction is made up. Let's take the case where the interval is divided up into 12 equal divisions. One division represents 1/12. Two divisions represent 2/12 which together represent 1/6 because 6*2=12 and the interval between 0 and 1 contains 6 sixths. Three divisions represent 3/12 which represent 1/4 because 4*3=12 and the interval between 0 and 1 contains 4 fourths or quarters, just like there are four quarters to $1. Four divisions represent 4/12 or one third. Five divisions just represent 5/12 and 7 divisions represent 7/12; but 6 divisions represent 6/12 or one half and two lots of these make 1. So 2*6/12 is the same as 2*1/2. 11 divisions make 11/12. Eight divisions is 8/12. Since 4/12 is a third, 8/12 must be two thirds because 2*4=8 or 2*4/12=8/12. Nine divisions are 3*3/12=9/12, and since 3/12 is the same as a quarter, 9/12 must be three quarters. Ten divisions make 10/12, but since 5*2/12=10/12 and 2/12 is one sixth, 10/12 must be 5 sixths or 5/6. If the interval 0 to 1 is divided into 60 divisions we get more fractions. The divisions help us to add and subtract. The number of divisions is associated with the least common multiple (LCM). So the LCM of 2, 3, 4 and 6 is 12. For 60 as LCM we have 2, 3, 4, 5, 6, 10, 12, 15, 20, 30. These numbers of divisions give us the fractions 1/30, 1/20, 1/15, 1/12, 1/10, 1/6, 1/5, 1/4, 1/3, 1/2. So to add 1/5 and 1/6 we use the divisions 12/60+10/60=22/60=11/30 because 22=2*11 and 22/60=2*11/60=11/30. We can also see that 11/30-1/5 is the same as 22/60-12/60=10/60=1/6. This should give you some idea how dividing 0 to 1 into different divisions helps you see how fractions add and subtract.
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Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42}

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42} Please just solve the set provided above!!!! This will be a bit more involved than the systems with two unknowns, but the process is the same. The plan of attack is to use equations one and two to eliminate z. That will leave an equation with x and y. Then, use equations one and three to eliminate z again, leaving another equation with x and y. Those two equations will be used to eliminate x, leaving us with the value of y. I'll number equations I intend to use later so you can refer back to them. That's enough discussion for now. 1)  3x-2y+2z=30 2)  -x+3y-4z=-33 3)  2x-4y+3z=42 Equation one; multiply by 2 so the z term has 4 as the coefficient. 3x - 2y + 2z = 30 2 * (3x - 2y + 2z) = 30 * 2 4)  6x - 4y + 4z = 60 Add equation two to equation four:   6x - 4y + 4z =  60 +(-x + 3y - 4z = -33) ----------------------   5x - y       = 27 5)  5x - y = 27 Multiply equation one by 3. Watch the coefficient of z. 3 * (3x - 2y + 2z) = 30 * 3 6)  9x - 6y + 6z = 90 Multiply equation three by 2. Again, watch the coefficient of z. 2 * (2x - 4y + 3z) = 42 * 2 7)  4x - 8y + 6z = 84 Subtract equation seven from equation six.   9x - 6y + 6z = 90 -(4x - 8y + 6z = 84) ----------------------   5x + 2y      =  6 8)  5x + 2y = 6 Subtract equation eight from equation five. Both equations have 5 as the coefficient of x. We eliminate x this way.   5x -  y = 27 -(5x + 2y = 6) ---------------       -3y = 21 -3y = 21 y = -7  <<<<<<<<<<<<<<<<<<< At this point, I am confident that I followed the correct procedures to arrive at the value for y. Use that value to determine the value of x. ~~~~~~~~~~~~~~~ Plug y into equation five to find x. 5x - y = 27 5x - (-7) = 27 5x + 7 = 27 5x = 27 - 7 5x = 20 x = 4  <<<<<<<<<<<<<<<<<<< Plug y into equation eight, too. 5x + 2y = 6 5x + 2(-7) = 6 5x - 14 = 6 5x = 6 + 14 5x = 20 x = 4    same value for x, confidence high Proceed, solving for the value of z. ~~~~~~~~~~~~~~~ Plug both x and y into equation one. We will solve for z. Equation one: 3x - 2y + 2z = 30 3(4) - 2(-7) + 2z = 30 12 + 14 + 2z = 30 26 + 2z = 30 2z = 30 - 26 2x = 4 z = 2  <<<<<<<<<<<<<<<<<<< Continue using the original equations to check the values. Equation two: -x + 3y - 4z = -33 -(4) + 3(-7) - 4z = -33 -4 - 21 - 4z = -33 -25 - 4z = -33 -4z = -33 + 25 -4z = -8 z = 2   same value for z, looking good Equation three: 2x - 4y + 3z = 42 2(4) - 4(-7) + 3z = 42 8 + 28 + 3z = 42 36 + 3z = 42 3z = 42 - 36 3z = 6 z = 2  satisfied with the results We have performed several checks along the way, thus proving all three of the values. x = 4, y = -7 and z = 2
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solve three way equation with z. 1:3x-2y+2z=30 2:-x+3y-4z=-33 3:2x-4y+3z=42

Solve {3x-2y+2z=30, -x+3y-4z=-33, 2x-4y+3z=42} Please just solve the set provided above!!!! 1)  3x-2y+2z=30 2)  -x+3y-4z=-33 3)  2x-4y+3z=42 Equation one; multiply by 2 so the z term has 4 as the coefficient. 3x - 2y + 2z = 30 2 * (3x - 2y + 2z) = 30 * 2 4)  6x - 4y + 4z = 60 Add equation two to equation four:   6x - 4y + 4z =  60 +(-x + 3y - 4z = -33) ----------------------   5x - y       = 27 5)  5x - y = 27 Multiply equation one by 3. Watch the coefficient of z. 3 * (3x - 2y + 2z) = 30 * 3 6)  9x - 6y + 6z = 90 Multiply equation three by 2. Again, watch the coefficient of z. 2 * (2x - 4y + 3z) = 42 * 2 7)  4x - 8y + 6z = 84 Subtract equation seven from equation six.   9x - 6y + 6z = 90 -(4x - 8y + 6z = 84) ----------------------   5x + 2y      =  6 8)  5x + 2y = 6 Subtract equation eight from equation five. Both equations have 5 as the coefficient of x. We eliminate x this way.   5x -  y = 27 -(5x + 2y = 6) ---------------       -3y = 21 -3y = 21 y = -7  <<<<<<<<<<<<<<<<<<< ~~~~~~~~~~~~~~~ Plug y into equation five to find x. 5x - y = 27 5x - (-7) = 27 5x + 7 = 27 5x = 27 - 7 5x = 20 x = 4  <<<<<<<<<<<<<<<<<<< Plug y into equation eight, too. 5x + 2y = 6 5x + 2(-7) = 6 5x - 14 = 6 5x = 6 + 14 5x = 20 x = 4    same value for x Proceed, solving for the value of z. ~~~~~~~~~~~~~~~ Plug both x and y into equation one. We will solve for z. Equation one: 3x - 2y + 2z = 30 3(4) - 2(-7) + 2z = 30 12 + 14 + 2z = 30 26 + 2z = 30 2z = 30 - 26 2x = 4 z = 2  <<<<<<<<<<<<<<<<<<< Continue using the original equations to check the values. Equation two: -x + 3y - 4z = -33 -(4) + 3(-7) - 4z = -33 -4 - 21 - 4z = -33 -25 - 4z = -33 -4z = -33 + 25 -4z = -8 z = 2   same value for z Equation three: 2x - 4y + 3z = 42 2(4) - 4(-7) + 3z = 42 8 + 28 + 3z = 42 36 + 3z = 42 3z = 42 - 36 3z = 6 z = 2  satisfied with the results   x = 4, y = -7 and z = 2
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3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer how do you solve it and what are the answers ? 1)  3x + 6y – 6z = 9 2)  2x – 5y + 4z = 6 3)  -x +16y + 14z = -3 The first objective is to eliminate z so we can solve for x and y. It's a multi-step process, so follow along. Multiply equation one by 4. 4 * (3x + 6y – 6z) = 9 * 4 4)  12x + 24y - 24z = 36 Multiply equation 2 by 6. 6 * (2x – 5y + 4z) = 6 * 6 5)  12x - 30y + 24z = 36 Now, we have two equations with a "24z" term. Add the equations and the z drops out. Add equation five to equation four.    12x + 24y - 24z = 36 +(12x - 30y + 24z = 36) ----------------------------------    24x -  6y           = 72 6)  24x - 6y = 72 The same process applies to equations two and three. Multiply equation two by 7 this time. 7 * (2x – 5y + 4z) = 6 * 7 7)  14x - 35y + 28z = 42 Multiply equation three by 2. 2 * (-x + 16y + 14z) = -3 * 2 8)  -2x + 32y + 28z = -6 Subtract equation eight from equation seven.   14x - 35y + 28z = 42 -(-2x + 32y + 28z = -6) ---------------------------------   16x - 67y          = 48 9)  16x - 67y = 48 Looking at equations six and nine, it would be simpler to eliminate the x. The multipliers are smaller. Multiply equation six by 2. 2 * (24x - 6y) = 72 * 2 10)  48x - 12y = 144 Multiply equation nine by 3. 3 * (16x - 67y) = 48 * 3 11)  48x - 201y = 144 Subtract equation eleven from equation 10.   48x -   12y = 144 -(48x - 201y = 144) ---------------------------          189y  =   0 189y = 0 y = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Substitute that value into equations six and nine to solve for x and verify. Six: 24x - 6y = 72 24x - 6(0) = 72 24x - 0 = 72 24x = 72 x = 3  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Nine: 16x - 67y = 48 16x - 67(0) = 48 16x - 0 = 48 16x = 48 x = 3    same answer for x To solve for z,substitute the x and y values into the three original equations. One: 3x + 6y – 6z = 9 3(3) + 6(0) – 6z = 9 9 + 0 - 6z = 9 9 - 6z = 9 -6z = 9 - 9 -6z = 0 z = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Two: 2x – 5y + 4z = 6 2(3) – 5(0) + 4z = 6 6 - 0 + 4z = 6 6 + 4z = 6 4z = 6 - 6 4z = 0 z = 0     same answer Three: -x +16y + 14z = -3 -(3) +16(0) + 14z = -3 -3 + 0 + 14z = -3 -3 + 14z = -3 14z = -3 + 3 14z = 0 z = 0     once again, same answer x = 3, y = 0, z = 0
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Write an equation for: Twelve times h equals one hundred and twenty minus thirty-six.

12×h=120-36 ........
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write in decimal of one hundred twenty-five million, three hundred fifteen thousand eighty-six

125, 315, 086 ..............................
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6x-6y-4z=-10 -5x+4y-z=-12 2x+3y-2z=9

6x-6y-4z=-10 -5x+4y-z=-12 2x+3y-2z=9 1)  6x - 6y - 4z = -10 2)  -5x + 4y - z = -12 3)  2x + 3y - 2z = 9 Multiply equation two by 4. 4 * (-5x + 4y - z) = -12 * 4 4)  -20x + 16y - 4z = -48 Subtract equation four from equation one.       6x -    6y - 4z = -10 -(-20x + 16y - 4z = -48) -----------------------------    26x - 22y      = 38 5)  26x - 22y = 38 Multiply equation two by 2. 2 * (-5x + 4y - z) = -12 * 2 6)  -10x + 8y - 2z = -24 Subtract equation six from equation three.       2x + 3y - 2z =    9 -(-10x + 8y - 2z = -24) ----------------------------    12x - 5y      = 33 7)  12x - 5y = 33 Multiply equation five by 5. 5 * (26x - 22y) = 38 * 5 8)  130x - 110y = 190 Multiply equation seven by 22. 22 * (12x - 5y) = 33 * 22 9)  264x - 110y = 726 Subtract equation eight from equation nine.   264x - 110y = 726) -(130x - 110y = 190) --------------------------   134x        = 536 134x = 536 x = 4  <<<<<<<<<<<<<<<<<<< Plug the value of x into equation seven. 12x - 5y = 33 12(4) - 5y = 33 48 - 5y = 33 -5y = -15 y = 3  <<<<<<<<<<<<<<<<<<< Plug the values for x and y into equation two. -5x + 4y - z = -12 -5(4) + 4(3) - z = -12 -20 + 12 - z = -12 -z = -12 + 20 - 12 -z = -4 z = 4  <<<<<<<<<<<<<<<<<<< x = 4, y = 3, z = 4 Check the answers by plugging all three values into the original equations. One: 6x - 6y - 4z = -10 6(4) - 6(3) - 4(4) = -10 24 - 18 - 16 = -10 6 - 16 = -10 -10 = -10 Two: -5x + 4y - z = -12 -5(4) + 4(3) - 4 = -12 -20 + 12 - 4 = -12 -8 - 4 = -12 -12 = -12 Three: 2x + 3y - 2z = 9 2(4) + 3(3) - 2(4) = 9 8 + 9 - 8 = 9 9 = 9 All values are correct.
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