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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 ...

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!

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 ...

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

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 ...

• 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, ...

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) ...

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 ...

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 ...

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...

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.

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.

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.

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.

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.

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).

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.