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What is x to the fourth minus one all divided by x to the third all equal to zero

(x^4-1)/x^3=0

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Basic Algebra/Introduction to Basic Algebra Ideas/Exponents ...


Basic Algebra/Introduction to Basic Algebra ... Three times three times three times three equal three to the fourth ... three to the negative third power equals one ...
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Math Forum - Ask Dr. Math


Operating on Negative Numbers Date: ... = 4 2) minus x minus = plus ... If we add one negative six together (-6 x 1), we get -6. If we add zero negative sixes ...
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What is x to the fourth power minus 20x plus 64 equals 0


What is X squred minus 5X plus 4 divided by X minus 1 ... Half of a third of x equals a fourth of y ... by x minus six equals Zero? Math Find all possible ...
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what is (x to the 3rd power + x to the 2nd power) divided by ...


... (x to the 3rd power + x to the 2nd power) divided by ... minus 5X plus 4 divided by X minus 1 (4^t)(8t ... power half divided by 16 to the power one fourth
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Accessible Accuplacer College Level Math Study Guide


Six X to the fourth power Y to the fourth power minus six X to the third power Y to the ... than zero. Question One: ... five minus eight divided by two ...
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Simplifying rational expressions: higher degree terms (video ...


Simplifying rational expressions: higher degree ... I don't see an x to the fourth minus one, ... so x cannot be equal to zero, x cannot be equal to plus or minus one.
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What is 12 x to the fourth minus 3 x to the second - Answers.com


Answers.com ® WikiAnswers ® Categories Science Math and Arithmetic Calculus What is 12 x to the fourth minus 3 x ... equal to zero and solve. (x ... minus 12 ...
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What is x to the fourth minus one all divided by x to the third all equal to zero

The bottom (x^3) can't make the fraction 0. Only the top (x^4 - 1) can make the fraction 0. x^4 - 1 = 0 x^4 = 1 x^2 = 1 x = +- 1 x = 1, -1 . (If you want to get fancy, x = 1, -1, i, -i)
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4 times one third x minus one fourth plus four thirds x divided by 3 plus one equals six

4((1/3)x - 1/4) + (4/3)x divided by 3 + 1 = 6 (4/3)x - 1 + (4/9)x + 1 = 6 (4/3)x + (4/9)x = 6 (12/9)x + (4/9)x = 6 (16/9)x = 6 x = 6 * (9 / 16) x = 27 / 8
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Creata a fraction story

FIVE LITTLE PIGS GO TO THE MALL Mama Pig gave her five little pigs seven and a half dollars between them to spend at the mall. It was a cold day, twenty-three Fahrenheit, minus five Celsius, or five degrees below freezing. Off they trotted at a quarter to three in the afternoon. "How far is it?" the youngest pig asked after a while. "One point seven five miles from home," said the eldest. "What does that mean?" asked the youngest. "Well," explained the eldest, "if we divide the distance into quarter miles, it's seven quarters." "How long will it take to get there?" asked the pig in the middle. Her twin sister replied, "It's five past three now, so that means we've taken twenty minutes to get here. Remember the milestone outside our house? There's another one here, so we've come just one mile and we've taken a third of an hour. [That means our speed must be three miles an hour.] "How much longer?" the youngest asked. "Three quarters of a mile to go," the next youngest started. "Yes," said the eldest, "we get the time by dividing distance by speed, so that means three quarters divided by three, which is one quarter of an hour [which is fifteen minutes]." ["So what time will we arrive?" the youngest asked. "About twenty past three," all the other pigs replied together.] "That's thirty-five minutes altogether," the eldest continued, "which means that - let me see - seven quarters divided by three is seven twelfths of an hour. One twelfth of an hour is five minutes [so seven twelfths is thirty-five minutes]. Yes, that's right." When they got to the mall, it had started to snow. Outside there was a big thermometer and a sign: "COME ON IN. IT'S WARMER INSIDE!" The eldest observed: "It shows temperature in Fahrenheit and Celsius. See, there's a scale on each side of the gauge. It's warmer now than it was when we left home. The scales are divided into tenths of a degree. It says twenty-seven Fahrenheit exactly, and, look, that's the same as minus two point eight Celsius," speaking directly to the youngest, "because the top of the liquid is about eight divisions between minus two and minus three. Our outside thermometer at home is digital, but this is analogue." The youngest stuttered: "What's 'digital' and 'analogue'?" "Well," began one of the twins, "your watch is analogue, because it has fingers that move round the watch face. Our thermometer at home is digital, because it just shows the temperature in numbers." ["Yes," said the eldest, "it would show thirty-seven point zero degrees Fahrenheit, four degrees warmer, and minus two point eight Celsius, two point two degrees warmer than when we left."] Continued in comment... 
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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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what is -t+5=t-19

People have a hang-up over negatives, but really there's no need to panic. After all if the temperature goes down to -2 it just means 2 degrees below zero. You can see this on a thermometer and everyone understands what it means, no problem. Just because negatives show up in mathematical and algebraic expressions is no big deal either. Back to your problem. To solve the equation and find t we need to gather together the unknowns (in this case the variable t) and the knowns, which are just numbers. We need to have the knowns on one side of the equals and the unknowns on the other side for this type of equation. It doesn't matter which side is used for what, although most people feel happier with the unknown(s) on the left and plain numbers on the right. We don't want loads of negatives so what we'll do is take the t's over to the right. When you cross from side of equals to the other, plus changes to minus and minus to plus. Divide changes to multiply and multiply changes to divide. Bring the -t from the left to right and it becomes +t or just t, added to the t already there makes 2t. Now bring the -19 from right to left where it becomes +19, added to the 5 already there makes 24. So we have 24=2t or put another way 2t=24. Take 2 from the left where it's multiplying over to the right where it divides into 24, making 12. So t=12. Bingo!
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| x +1| less than or equal to | x - 3 | solve . please help

| x + 1 | <= | x - 3 | With two absolute values, there are 4 possible solutions: First: x + 1 <= x - 3 Second: x + 1 <= -(x - 3) Third: -(x + 1) <= x - 3 Fourth: -(x + 1) <= -(x - 3) Let's work them out one at a time: . First: x + 1 <= x - 3 subtract x from both sides 1 <= -3 Never true.  This one can never happen. . Second: x + 1 <= -(x - 3) x +1 <= -x + 3 add x to both sides 2x + 1 <= 3 subtract 1 from both sides 2x < = 2 Answer:  x < = 1 . Third: -(x + 1) <= x - 3 -x - 1 <= x - 3 add x to both sides -1 <= 2x - 3 add 3 to both sides 2 <= 2x divide by 2 on both sides 1 <= x Answer:  x >= 1 . Fourth: -(x + 1) <= -(x - 3) -x - 1 <= -x + 3 add x to both sides -1 <= 3 Never true.  This solution can't happen. . So our 4 solutions are: Can't happen, x <= 1, x >= 1, can't happen. Either of the two solutions (the ones that can happen) can be true and the original equation is true. But our two solutions (x <= 1, x >=1) cover everything- there is no value for x that doesn't fit into one of those solutions.  That means x can be anything. Answer:  x can be any number.
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If you had 4 peices of cake and you want to share 2/3 of it with a friend, how much pieces will you have left?

If you want to share 2/3 of 4 pieces of cake you will have to find a way of dividing the pieces because 2/3 of 4 is 8/3=2 2/3. You then need to halve 2 2/3 to share with your friend, leaving you with 1 1/3 pieces. But you have a third of the 4 pieces left because you only shared 2/3 of 4 pieces. A third of 4 pieces is 4/3=1 1/3 pieces (plus what you shared with your friend, which was also 1 1/3 pieces, making 2 2/3 pieces). Let's make the problem a lot easier. Divide each of the 4 pieces of cake into three equal parts, so you will have 12 smaller pieces of cake. Now it's much easier to divide. 2/3 of 12 small pieces is 8 small pieces. When you share 8 of these small pieces with your friend, you will have 4 small pieces each. So you have 4 small pieces by sharing and another 4 small pieces you didn't share. So you have 8 small pieces altogether. Let's suppose your cake is round. Now imagine it's the face of a clock. If the whole cake is divided into 4 pieces then it's like cutting the cake so that one piece is the shape of the part of the clock between 12 and 3; the second piece between 3 and 6; the third piece between 6 and 9; and the fourth piece between 9 and 12. But each piece can be cut into 3. For example, the first piece would be like cutting a shape like 12 to 1, then 1 to 2, and 2 to 3. The part of the cake you will be sharing is between 12 and 8 going clockwise 12, 1, 2, 3, 4, 5, 6, 7, 8 that is two quarters (1 to 3 and 3 to 6) and 2/3 of another quarter 7 to 8, making 2 2/3). When you share with a friend, you have to divide these 8 segments so that you have 4 each. I hope you can see how this works.
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find all real zeros of f(x)=6x^3-5x-1

Since 1 is the constant and the only roots of 1 have a magnitude of 1, we can see that ax=1 or ax=-1 is a root, where a is a constant. If we put x=1 into the function, it does in fact equal zero. So x-1 is a factor, which we can divide by, using synthetic division: 6x^2+6x+1. Therefore f(x)=(x-1)(6x^2+6x+1). The quadratic implies that we have to look at the factors of 6 (that is 6 and 1, and 2 and 3) and the factors of 1 (that is 1 and 1, and -1 and -1) and combine these possibilities to get the coefficient of the middle term, which is 6. We also know that from the sign of the constant we have to add the products of the prospective factors, and we know from that that we can have no minus signs. So we're looking for 1 and 1 as the constants, and the products are 2*1+3*1=5 or 6*1+1*1=7. Neither of these gives us the 5 we need. The solution of the quadratic is x=(-6+sqrt(36-24))/12=(-3+sqrt(3))/6=-0.2113 or -0.7887. Therefore the three roots are 1, -0.2113, -0.7887 approx.
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how to solve -4/9(2x-4)=48

Is this -4/(9(2x-4))=48? Or (-4/9)(2x-4)=48? We can find the answers to both questions. Let's take the first interpretation first and see what we get. To solve it we need to get rid of the fraction. In all questions of this type we take the denominator of the fraction (the bottom bit), and multiply both sides of the equation by it, which is the same as taking it across the equals sign from left, where it was a divisor over to the right, where it becomes a multiplier. So we get -4=48*9(2x-4). We have the unknown x inside the brackets, so we have to get it out into the open by multiplying what's outside by everything in the brackets. That gives us -4=864x-1728. Already this looks like the wrong interpretation of the question because the numbers are big and clumsy. But we'll carry on. The next thing to do is get the knowns (ordinary numbers) on one side of the equals and the unknowns on the other. So -1728 on the right becomes +1728 when we take it across to the left (because the equals sign changes + to - and - to +, multiply to divide and divide to multiply). When we add 1728 to -4 it's the same as subtracting 4 from 1728, so we get 1724=864x. So we take 864 from right to left (changing multiply to divide as we move it) and we get 1724/864=x, which gives us a horrible 431/216 or 1.995 approx. seems unlikely then that this was the intended question. Let's go for the other interpretation: (-4/9)(2x-4)=48. One thing to notice (which also applied to the first interpretation) is that  the equation can be simplified by dividing both sides by 4. So (-1/9)(2x-4)=12. The only fraction we have is -1/9, so we just need to take the 9 across to the right where it multiplies 12. Therefore -(2x-4)=108. We can simplify this because 2 divides into both sides of the equation. So we get -(x-2)=54.  We can take the brackets with the minus sign over to the other side of the equation where the minus will become plus. So we get 0=54+x-2. Now we separate the knowns from the unknowns. -54+2=x. So x=-52. This looks more like the correct interpretation.
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y=1-2x y=x2-3x-5

  y=1-2x y=x2-3x-5  i assume its y=(x^2)-3x-5 where x^2 means x squared. Not sure what level you are at, but as both y's are the same, set them equal to one another, ie 1-2x=(x^2)-3x-5 then set them equal to zero ie 0=(x^2)-x-4  Then you can use the quadratic formula (if you know what that is) or you can factorise or complete the square.  Note- will not factorise properly so quad formula or completing square needed.  When using formula ->        a=1, b=-1, c=-4 1(+or-) sqrt 1^2 - 4*1*-4, all divided by 2 1 plus root 17 all over 2 or 1 minus root 17 all over 2 :)
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