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Wiki User
∙ 9y ago
Wiki User
∙ 9y agoIf a and b are integers, then a x b,would be ab but a plus b won't be ab.
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Find two consecutive integers whose sum is 72?
There are no two consecutive integers that add up to 72. You can prove it this way: Let our numbers be represented by the variables "a" and "b". We are told: a = b + 1 a + b = 72 So we can use substitution to solve for either variable: (b + 1) + b = 72 2b + 1 = 72 2b = 71 b = 35.5 But 35.5 is not an integer, so this condition can not be met.
If a plus b plus c not equal to 0 then a divided by b plus c equals b divided by c plus a equals c divided by a plus b prove that a equals b equals c?
Because there is no way to define the divisors, the equations cannot be evaluated.
What are the positive integers for which a plus b plus c plus d equals 20 and abcd equals 81?
If a, b, c, and d are positive integers that add up to 20 they each must be less than 20. If abcd=81 then a, b, c, and d must be factors of 81. The factors of 81 are 1,3,9 then 27, so there is no solution to this problem.
If A B and C are counting numbers and both A and B are multiples of C what can you say about A plus B?
A + B is also a multiple of C. ------------------------------------------- let k, m and n be integers. Then: A = nC as A is a multiple of C B = mC as B is a multiple of C → A + B = nC + mC = (n + m)C = kC where k = n + m kC is a multiple of C. Thus A + B is a multiple of C.
Are all negative rational numbers integers?
No, they are not because fractions can be negative also. fractions aren't integers
Prove that if a and b are rational numbers then a plus b is a rational number?
Because a is rational, there exist integers m and n such that a=m/n. Because b is rational, there exist integers p and q such that b=p/q. Consider a+b. a+b=(m/n)+(p/q)=(mq/nq)+(pn/mq)=(mq+pn)/(nq). (mq+pn) is an integer because the product of two integers is an integer, and the sum of two integers is an integer. nq is an integer since the product of two integers is an integer. Because a+b equals the quotient of two integers, a+b is rational.
How do you prove if a is less than b then a plus c is less than b plus c?
I believe that's usually treated as an axiom, meaning you don't prove it.
Find two consecutive integers whose sum is 72?
There are no two consecutive integers that add up to 72. You can prove it this way: Let our numbers be represented by the variables "a" and "b". We are told: a = b + 1 a + b = 72 So we can use substitution to solve for either variable: (b + 1) + b = 72 2b + 1 = 72 2b = 71 b = 35.5 But 35.5 is not an integer, so this condition can not be met.
What procedures will find the sum a plus b of two numbers where a and b represent any integers?
addition a + b
Prove that A intersect with B is the subset of A?
Let x be in A intersect B. Then x is in A and x is in B. Then x is in A.
What are the conditions to swap in c plus plus without temporary?
You can swap two integers without temporary storage by bitwise exclusive-or'ing them in a specific sequence...a ^= b;b ^= a;a ^= b;
If a and b integers and a b then a b?
If a and b are integers, then a times b is an integer.
Can a trinomial x2 plus bx plus c where b and c are integers be factored with integer coefficients if its discriminant is 35?
No.
If a plus b plus c not equal to 0 then a divided by b plus c equals b divided by c plus a equals c divided by a plus b prove that a equals b equals c?
Because there is no way to define the divisors, the equations cannot be evaluated.
How many ordered triples of positive integers satisfy the equation a plus b plus c equals 6?
There are 6 such triples.
What are the positive integers for which a plus b plus c plus d equals 20 and abcd equals 81?
If a, b, c, and d are positive integers that add up to 20 they each must be less than 20. If abcd=81 then a, b, c, and d must be factors of 81. The factors of 81 are 1,3,9 then 27, so there is no solution to this problem.
If AA plus bb plus cc equals dwhere a and b are consecutive positive integers and c equals ab then root d is what?
+/- 11
