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If x is an integer divisible by 3 then is x2 divisible by 3?
Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
An integer minus three times its reciprocal equals twenty six thirds What is the integer?
Call the unknown integer x. Then, from the problem statement, x - 3/x = 26/3, or:x2 - 3 = 26x/3; or x2 - (26/3)x - 3 = 0x = 9
What is the product of all integer values ofx for which X2-9 is a prime number?
The answer is -16.
The square of an integer is equal to three times itself minus 2 write the statement in equation form and determine the integers?
x2 = 3x - 2 x2 - 3x + 2 = 0 (x-1)(x-2) = 0 x=1;x=2
What whole number is divisible by 3037?
any multiple of 3037 such at itself or 3037 x2 etc.
What is a counter example to falsify If x is an integer divisible by 3 then x2 is an integer divisible by 3?
Suppose x = sqrt(3*a) where a is an integer that is not divisible by 3. then x2 = 3*a which is divisible by 3. but x is not even rational and so is not an integer and is certainly not divisible by 3.
If x is an integer divisible by 3 then is x2 divisible by 3?
Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
Find number of integer solution of X1 x2 x324?
find number of integer solution of X1+x2+x3=24
If the domain of f(x)-x2 is integer values of x such that -3-x?
5
If x is a positive integer solve x2 6x 16?
Um.......is it x2+6x or x2-6x. I think your missing some things here champ
If x2 is an integer divisible by 3 then x is divisible by 3?
Yes. If x is not divisible by 3 then x leaves a remainder of 1 or 2 when it is divided by 3. That is, x is of the form 3y+z where z = 1 or 2. Then x2 = (3y+z)2 = 9y2 + 6yz + z2 = 3(3y2 + 2yz) + z2 The first part of this expression is clearly a multiple of 3, but z2 is not. Whether z = 1 or 2, z2 leaves a remainder of 1 when divided by 3.
An integer minus three times its reciprocal equals twenty six thirds What is the integer?
Call the unknown integer x. Then, from the problem statement, x - 3/x = 26/3, or:x2 - 3 = 26x/3; or x2 - (26/3)x - 3 = 0x = 9
What is The square of a positive integer is 4 more than 3 times the integer?
x2=3x+4 x2-3x-4=0 (x-4)(x+1)=0 x=4 or x=-1 Since you specified a positive integer, the number in question is 4.
6 times an integer minus 5 equals the square of the integer. find the integer?
If we write the problem as 6x-5=x2, then we can write in the form of ax2+bx+c: -x2+6x-5. Then we can use the quadratic equation, x = (-b ± √(b2 - 4ac))/2a, and put in our own values to get the equation x = 3 ± 2. Therefore, x1= 1 and x2=5.
Is the sum of the squares of integers separated by a value of 2 will always be even?
It looks to me as if that's true. I reasoned thusly, and scratched it outon the margin of my coffee-stained notepad:You gave me integers separated by 2, so the integers are [x] and [x+2].-- Their squares are [x2] and (x+2)2-- That's [x2] and [x2 + 4x + 4].-- The sum of their squares is [x2 + x2 + 4x + 4]= [ 2x2 + 4x + 4 ]-- Since [x] is an integer, each term in that trinomial is an integer.-- The coefficients are '2' and '4', so each term is an even number.-- So their sum is even.-- Q.E.D.
What is the product of all integer values ofx for which X2-9 is a prime number?
The answer is -16.
The square of an integer is equal to three times itself minus 2 write the statement in equation form and determine the integers?
x2 = 3x - 2 x2 - 3x + 2 = 0 (x-1)(x-2) = 0 x=1;x=2
