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What is 0.6x + 4.8=7.2 in integers

0.6x + 4.8 =7.2

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Integers And Equations. 2(-3)(5)-20 - (-5) 15 ... 0. 2. 3. x + 1, x + 2. x+2, x + 4, x ... Multiplication and Division Equations with Integers-3x = -15. 20 = -4x-22 ...
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Whole Numbers and Integers - Math Is Fun


Whole Numbers and Integers Whole Numbers. Whole Numbers are simply the numbers 0, 1, 2, 3, 4, 5, ... (and so on) No Fractions!
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What is the total number of pairs of integers (x,y) which ...


What is the total number of pairs of integers (x,y) which satisfy the equation: x^2-4xy+5y^2+2y-4=0? ... x y [1,] -7 -3 [2,] -6 -2 [3,] -5 -3 [4,] -2 -2 [5,] -2 0 [6 ...
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How to find the number of integers that satisfy the ...


How can I find the number of integers that satisfy the ... Integers in between are [math]\left\{1,2,3,4,5\right ... What are the roots of [math]x^3+5x^2+6x+7=0

  • Integers

    www.themathleague.com/index.php/about-the-math-league/...

    9 > 4, 6 > -9, -2 > -8, and 0 > -5. ... In the following examples, we convert the subtracted integer to its opposite, and add the two integers. 7 - 4 = 7 + (-4) = 3

  • Integer - Wikipedia

    en.wikipedia.org/wiki/Integer

    For example, 21, 4, 0, and −2048 are integers, while 9.75, ... (−4,8) (−3,6) (−2,4) (−1,2) (0,0) (1,1) (2,3) (3,5) …} If N = {1, 2, 3, ...} then consider ...
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  • Intro 2 Integers - Math Goodies


    The Integers are introduced. ... We add the distance from + 20,320 to 0, ... Jenny has $2. She earns $5, spends $10, earns $4, ...
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    Subtracting Positive and Negative Integers


    Subtracting Negative and Positive Integers. To subtract integers, change the sign on the integer that is to be subtracted. ... 6 - (+2) = 6 - 2 = 4 Example: -6 - (-2) ...
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    1. Integers - IntMath - Interactive Mathematics - Learn math ...


    1. Integers; 1a. Counting Game; 2 ... 3 + −7 = −4. When we add 2 integers, we get an integer. ... Complete the magic square, using only the integers: −10, −8 ...
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    Write x^2 - 6x + 7 = 0 in the form (x - a)^2 = b, where a and ...


    Write x^2 - 6x + 7 = 0 in the form (x - a)^2 = b, ... ^2 = b, where a and b are integers. 2. Ask for details ; Follow; Report; by Bestha1n0irelianado 07/25/2016. Log ...
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    x^3+y^3=z^3?

      x^3+y^3=z^3⇒(x+y)(x^2-xy+y^2)=z^3. If x+y=z, then x^2-xy+y^2=z^2; but that implies x^2-xy+y^2=x^2+2xy+y^2, and  3xy=0, which is the trivial solution and is disallowed by the theorem. So x+y needs to be an integer factor of z, x+y=az and x^2-xy+y^2=z^2/a; or x+y=az^2 and the quadratic x^2-xy+y^2=z/a. This respectively implies that (1) x^2-xy+y^2=(x^2+2xy+y^2)/a^2; or (2) x^2-xy+y^2=sqrt((x+y)/a)/a. (1) The equation becomes x^2(a^2-1)-xy(a^2+2)+y^2(a^2-1)=0 for a>1, and, dividing through by a^2-1 we get: x^2-xy(a^2+2)/(a^2-1)+y^2=0. Solve this by completing the square: (x-y(a^2+2)/(2(a^2-1)))^2+y^2-y^2((a^2+2)/(2(a^2-1))^2=0. x-y(a^2+2)/(2(a^2-1))=+ysqrt((a^2+2)^2-4(a^2-1)^2)/(2(a^2-1))= +ysqrt((a^2+2+2a^2-2)(a^2+2-2a^2+2))/(2(a^2-1))=+ysqrt(3a^2(4-a^2))/(2(a^2-1))= +aysqrt(3(4-a^2))/(2(a^2-1)). The square root cannot be negative and a>1, so a=2 and x=y(a^2+2)/(2(a^2-1))=6y/6=y. But this solution is based on the initial assumption that the quadratic could be equal to z/a. Clearly x^3+y^3=2x^3 cannot be z^3 where z is an integer. So the assumption (supposed condition) is false and there are no solutions under (1). (2) If x^2-xy+y^2=z/a and we write z/a=k, where k is a positive integer, then we can rewrite the quadratic: (x-y/2)^2-y^2/4+y^2=k, so (x-y/2)^2=k-(3/4)y^2. If y is an even integer y=2w where w is an integer. Then (x-w)^2=k-3w^2. So (x-w)=sqrt(k-3w^2). The left-hand side is an integer so the right-hand side must also be an integer, and this can only happen if k-3w^2=p^2 where p is an integer >0. So k=p^2+3w^2. x=w+p; y=2w; z=a(p^2+3w^2). We have the three variables in terms of three parameters, a, p and w. For the special case of p=0, we can see the effect of substituting the parametrised values into the quadratic: w^2-2w^2+4w^2=3w^2 and into x+y=3w, the product of which is 9w^3=x^3+y^3. This sum cannot be a perfect cube because 9 is not a perfect cube, therefore there are no solutions to the cubic equation when p=0. When p>0, x+y=3w+p and the quadratic is w^2+2pw+p^2-2w^2T2wp+4w^2=3w^2+p^2. The product is (3w+p)(3w^2+p^2)=9w^3+3wp^2+3pw^2+p^3=(2w)^3+(p+w)^3=y^3+x^3. If y is odd y=2v-1 where integer v>0 and (x-v+(1/2))^2=k-(3/4)(4v^2-4v+1)=k-3v(v-1)-3/4; (x-v)^2+x-v+1=k-3v(v-1) More...
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    Proof by the method of contradiction ,if sin a =O then a =k(pi) for any integer k

    Assume a≠kπ as the general solution of sin(a)=0. If k is any integer and zero is an integer then k=0 is a possible value for k. But if k=0 then, since a≠kπ on our assumption, a≠0 is a solution for sin(a)=0. In a right-angled triangle sine=opposite side/hypotenuse by definition. So if sin(a)=0 the opposite side length must be permitted to be zero, since no value of the hypotenuse would make the quotient zero. As the length of the opposite side approaches zero, the angle, a, gets smaller. In the limit, the opposite side length becomes zero, and the angle a must also be zero, so a=0 is a solution to sin(a)=0, so contradicting the implication that a≠0. This would eliminate k=0 as an integer. Also, sin(π-a)=sin(a). When sin(a)=0, sin(π)=0. When k=1, if we assume a≠π is a solution we have the contradiction that sin(π)=0. And because sine is a periodic function sin(2π+a)=sin(a) so the argument can be extended to sin(2π), sin(2π+π)=sin(3π)= etc. Therefore we are led to the conclusion that a=kπ is a solution when sin(a)=0.  
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    Pete challenges his friend Jill to find two consecutive odd integers that have the following relationship.

    Two consecutive odd integers can be represented by 2x-1 and 2x+1. Their product is 4x^2-1 and their sum is 4x. So, 4x^2-1=3(4x+6)=12x+18. This is the quadratic: 4x^2-12x-19=0. This has no rational solutions. Or, reading the question slightly differently: 4x^2-1=12x+6, so 4x^2-12x-7=0. In the second case, we can factorise: (2x-7)(2x+1)=0, so x=3.5 or -0.5. The two integers are: 6 and 8, or -2 and 0. None of these are odd, so it isn't possible to find two consecutive odd integers. Why did I think there were two interpretations of the question? "3 times the sum of the integers plus 6" is ambiguous:  is it 3(sum+6) or 3*sum+6? Judging by the result, it was the latter.
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    the product of...

    The product of two consecutive integers is 306. Find the integers. The hardest part is understanding what the question is asking and setting up the problem.  Once you have this, it is a piece of cake.  Here's what we do: First what its asking: (Break the question down into segments) The product: Ok this means Multiply. Two consecutive integers: This means that the second number comes right after the first number in a number line. Ex. 1,2,3,4,5... The integers belong to the number 306: Factor 306.                    306                      ^                     2*153                          ^                         9*17                         ^                        3*3 Now look 3 * 3 * 2 = 18 The other factor 306 = 17 These two are consecutive numbers so these are your integers since 18 * 17 = 306. A more Algebraic approach would be to say: two consecutive intergers are multiplied together to make 306 x(x + 1) = 306 x^2 + x = 306 x^2 + x - 306 = 0 (x - 17)(x+18) = 0 x - 17 = 0; x = 17 x - 18 = 0; x = 18 There are your two consecutive integers. 
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    What is the diameter of a spiral coil of .65265 inch diameter pipe 100 feet long?

    The equation of a spiral in polar coordinates has the general form r=A+Bø, where A is the starting radius of the spiral and B is a factor governing the growth of the spiral outwards. For example, if B=0, there is no outward growth and we just have a circle of radius A. A horizontal line length A represents the initial r, and the angle ø is the angle between r and this horizontal line. So r increases in length as ø increases (this angle is measured in radians where 2(pi) radians = 360 degrees, so 1 radian is 180/(pi)=57.3 degrees approximately.) If B=1/2 and A=5", for example, the minimum radius would be 5" when ø=0. When ø=2(pi) (360 degrees), r=5+(pi), or about 8.14". This angle would bring r back to the horizontal position, but it would be 8.14" instead of the initial 5". At ø=720 degrees, the horizontal line would increase by a further 3.14". Everywhere on the spiral the spiral arms would be 3.14" apart. What would B be if the spiral arms were 0.65625" apart? 2(pi)B=0.65625, so B=0.65625/(2(pi))=0.10445". The equation of the spiral is r=5+0.10445ø. To calculate the length of the spiral we have two possible ways: an approximate value based on the similarity between concentric circles and a spiral; or an accurate value obtainable through calculus. The approximate way is to add together the circumferences of the concentric circles: L=2(pi)(5+(5+0.65625)+...+(5+0.65625N)) where L=spiral length and N is the number of turns. L=2(pi)(5N+0.65625S) where S=0+1+2+3+...+(N-1)=N(N-1)/2. This formula arises from the fact that the first and last terms (0, N-1) the second and penultimate terms (1, N-2) and so on add up to N-1. So, for example, if N were 10 we would have (0+9)+(1+8)+(2+7)+(3+6)+(4+5)=5*9=45=10*9/2. If N were 5 we would have 0+1+2+3+4=10=(0+4)+(1+3)+2=5*4/2. L=12*100 inches. L=1200=2(pi)(5N+0.65625N(N-1)/2)=(pi)N(10+0.65625(N-1))=(pi)N(9.34375+0.65625N). If the external radius is r1 and the internal radius is r then the thickness of the spiral is r1-r and since 0.65625 is the gap between the spiral arms N=(r1-r)/0.65625. N is an integer, but, since it is unlikely that this equation would actually produce an integer we would settle for the nearest integer. If we solve this equation for N, we can deduce the external radius and diameter of the spiral: N(9.34375+0.65625N)=1200/(pi)=381.97; 0.65625N^2+9.34375N-381.97=0 and N=(-9.34375+sqrt(1089.98))/1.3125=18 (nearest integer). This means that there are 18 turns of the spiral to make the total length about 100 feet. If X is the final external diameter of the coiled pipe and the internal radius is 5" (the minimum allowable) then X/2 is the external radius, so N=((X/2)-5)/0.65625. We found N=18 so we can find X: X=2*(0.65625*18+5)=33.625in. Solution using calculus Using calculus, we can work out the relationship between the length of the spiral and other parameters. We start with any polar equation r(ø) and a picture: draw a line representing a general value of r. At a small angle dø to this line we draw another line a little bit longer, length r+dr. Now we join the ends together to make a narrow-angled triangle AOB where angle AOB=dø and AB=ds, the small section of the curve. In the triangle AO is length r and BO is length r+dr. If we mark the point C along BO so that CO is length r, the same as AO, we have an isosceles triangle COA. Because the apex angle is small, CA=rdø, the length of the arc of the sector. In triangle ABC, CB=dr, AB=ds and CA=rdø. By Pythagoras, AB^2=CB^2+CA^2, that is, ds^2=dr^2+r^2dø^2, because angle BCA is a right angle as dø tends to zero. The length of the curve is the result of adding the tiny ds values together between limits of r or ø. We can write ds=sqrt(dr^2+r^2dø^2). If we divide both sides by dr, we get ds/dr=sqrt(1+(rdø/dr)^2) so s=integral(sqrt(1+(rdø/dr)^2)dr, where s is the length of the curve. The integral is definite if we define the limits of r. For our spiral we have r=A+Bø, making ø=(r-A)/B and B=p/(2(pi)), where p is the diameter of the pipe=0.65625", so we can substitute for ø in the integral and the limits for r are A to X/2, where A is the inner radius (A=5") and X/2 is the outer radius. dø/dr=2(pi)/p, a constant=9.57 approx. s=integral(sqrt(1+(2(pi)r/p)^2)dr) between limits r=A to X/2. After the integral is calculated, we solve for X putting s=1200". The expression (2(pi)r/p)^2 is large compared to 1, so s=integral((2(pi)r/p)dr) approximately and s=[(pi)r^2/p] (r=A to X/2); therefore, since we know s=1200, we can write ((pi)/p)(X^2/4-A^2)=1200. Therefore X=2sqrt(1200p/(pi))+A^2)=33.21". Compare this answer with the one we got before and we can see they are close. [We could get a formal solution to the integral, using hyperbolic trigonometric or other logarithmic functions, but such a solution would make it very difficult or tedious to solve for X, since X would appear in logarithmic expressions and in other expressions making it difficult or impossible to isolate X. For example, the next term in the expansion of the integral would be (p/(4(pi))ln(X/2A), having a value of about 0.06. It is anticipated, therefore, that an approximation would be sufficient in this problem with the given figures.] We can feel justified in using the formula for finding the length of pipe, L, when X=6'=72": L=((pi)/p)(1296-25)=6084.52"=507' approximately. This length of pipe would hold 507/100*0.96 gallons=4.87 gallons.      
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    find the number of positive integers which satisfies the inequality

    e^x(x-2)(-x+1)(x+1)²/(x+1)(x²+ x+1)>=0 factor out an x+1 e^x(x-2)(-x+1)(x+1)/(x²+ x+1)>=0 e^x is always positive, regardless of the value of x, so divide e^x out (x-2)(-x+1)(x+1)/(x²+ x+1)>=0 since we're restricting our possible answers to positive integers, x^2 + x + 1 is always positive, regardless of the value of x, so multiply x^2 + x + 1 out (x-2)(-x+1)(x+1)>=0 x - 2 is >=0 when x>=2 -x+1 is <=0 when x>=1 x+1 is >=0 when x>=1  (remember, x can't = -1 or 0) x = 0  gives us - + +, a negative result, isn't >= 0 x = 1 gives us - 0 +, a zero result, is >= 0 x = 2 gives us 0 - +, a zero result, is >= 0 x = 3 gives us + - +, a negative result, isn't >= 0 Note:  Any value for x above 2 won't change the sign of the original equation. Answer:  There are 2 positive integers (1 and 2) that satisfy the inequality e^x(x-2)(-x+1)(x+1)²/(x+1)(x²+ x+1)>=0.
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    how do I evaluate (cos(pix/2))

    cos((pi)x/2) can be evaluated easily for integer x. Let x=4n, where n is an integer, so we get cos(2n(pi))=1. If x=4n-1 we get (pi)x/2=2n(pi)-(pi)/2, so cos((pi)x/2)=cos(2n(pi))cos(-(pi)/2)+sin(2n(pi))sin((pi)/2)=1*0+0*1=0. If x=4n-2, the expression = -1; if x=4n-3, the expression = 0. So we can plot the expression's zeroes, maxima and minima, and then we can look at non-integer values of x. When x=0, 1, 2, 3, ... we have 1, 0, -1, 0, 1, 0, -1, 0, ... as outputs. In between we have a typical sine wave pattern oscillating between the extrema and cutting the axis as it dips and dives all the way from minus infinity to plus infinity. You can either draw a graph to evaluate the expression or build a table. Because x is a variable there is no one single evaluation so you need to use a calculator to put in various values for x or refer to basic tables. You don't need to build a table or graph for every value of x because the expression values repeat: x=0 to 4 is the only range of x you need, because x=-8 to -4, -4 to 0, 0 to 4, 4 to 8, 8 to 12, etc., give you the same values of the expression.
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    What are four integers between 0 and +10 that does not model a function?

    See this page ( http://www.algebra.com/algebra/homework/Functions/Functions.faq.question.393212.html ) for info on ordered pairs that do not model a function. So ordered pairs are things like (1,2), (3,4), and so on. Ordered pairs that model a function have to have the first coordinates (the 'x' in (x,y) ) all different.  The second coordinates (the 'y' in (x,y) ) can have values that are the same. For this problem we want four ordered pairs, so let's try: (0,-5), (1,-6), (2,-7), (3,-8) Those four ordered pairs do model a function because the first coordinates (0,1,2,3) are all different. We want four ordered pairs that do *not* model a function, so there has to be at least one match between the first coordinates.  We could do this: (0,-5), (0,-6), (2,-7), (3,-8) There are many possible solutions, but this one works because the first coordinates (0,0,2,3) have a repeat (two zeroes).
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    how many ways are there to add and get the sum of 180

    There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!
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    k^2-k-20/18

    I read this as (k^2-k-20)/18. What pair of factors of 20 differ by 1? This is essentially what the quadratic in k is asking. The pairs of factors of 20 are (1,20), (2,10), (4,5) and the only pair with a difference of 1 is (4,5). The larger of the two has a minus and the other a plus to make the difference -1 which is the coefficient of the middle term. So k^2-k-20 factorises: (k-5)(k+4). And we place this over 18: (k-5)(k+4)/18. This won't factorise any further. When I was reviewing this question I felt there was more to it. Supposing the question was: what values can k take so that the expression is an integer? I spotted something interesting. The zeroes of the quadratic are 5 and -4: either of these values make the expression zero. But 5+4=9, and 9 is a factor of 18, the denominator. We can write the factors of the quadratic as y(y-9)/18, where y replaces k+4; or we can write y(y+9)/18, where y replaces k-5. The pairs of factors of 18 are (1,18), (2,9), (3,6). If the numerator contains any of these pairs, the expression will be an integer, or whole number, positive or negative. What values of y must we have so that y(y-9)/18 is an integer? Let's start with y=0; the expression is zero, which is an integer. Now y=3; we have 3(-6)/18=-1, another integer. y=6; 6(-3)/18=-1. y=9; we get 0. y=12 we get 12*3/18=2. y=15; we get 15*6/18=5. In fact, all multiples of 3 work, so we can write y=3N as the general solution, where N is a positive or negative integer. What about k? We know that y=k+4 or k-5 so k+4=3N or k-5=3N are solutions. We can write these solutions as k=3N-4 or k=3N+5. If N=0 we have -4 and 5, which are the zeroes of the quadratic. If N=1, k=-1 or 8; if N=2, k=2 or 11; if N=3, k=5 or 14; if N=4, k=8 or 17; and so on. If you substitute these and other values of k (according to the formula) you'll see that the expression is always an integer.
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