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What is the expression of The product of a number and 2 is 3 less than the quotient of a number and 2.?
2m = n/2 - 3 where m and n are the two numbers.
What 2 numbers with a sum of 12 and a product of 35?
Let the numbers by 'm' & 'n' Hence m + n = 12 mn = 35 Hence m = 35/n Substitute 35/n + n = 12 Multiply through by 'n' 35 + n^2 = 12n n^2 - 12n + 35 = 0 It is now in Quadratic Form ; Factor ( n - 7)(n - 5) = 0 Hence n= 7 or n= 5 It follows that m = 5 , or m = 7
What are your two numbers if their sum is -5 and their product is 4?
Let the number be 'm' & 'n' Hence Sum = m + n = -5 Product = mn = 4 Algebraically rearrange the product to m = 4/n Substitute into the Sum (4/n) + n = -5 Multiply through by 'n' Hence 4 + n^2 = -5n n^2 + 5n + 4 = 0 It is not in Quadrtatic form . to solve. Hence (n - 1)(n - 4) = 0 n = -1 & n = -4 Are the two numbers.
Product of 5m plus half of n?
5(m+/n
When a trinomial is factored as (x m)(x n) what is the product of m and n?
It is the constant term of the trinomial.
What is the product of 5 and m plus half of n?
If we are to find the product of 5 and m and n/2 (which is half of n), we have: 5 times m times n/2 = 5 x m x n/2 = 5mn/2
Where m and n are statements m n is called the of m and n?
Where m and n are statements m n is called the _____ of m and n.
If m and n are prime numbers such that the sum of m and n is greater than 20 and the product of m and n is less than 50 What is a possible value of the product of m and n?
38 or 46
What is the expression of The product of a number and 2 is 3 less than the quotient of a number and 2.?
2m = n/2 - 3 where m and n are the two numbers.
Two numbers have a sum of -9 and a product of -90 What are the numbers?
Let the two numbers be 'm' & 'n' Sum = m + n = -9 Product is mn = -90 Hence m = -90/n Substitute -90/n + n = -9 Multiply through by 'n' -90 + n^2 = - 9n n^2 + 9n - 90 = 0 It is now in quadratic form ; factor (n + 15)(n - 6) = 0 n = -15 & n = 6
What 2 numbers with a sum of 12 and a product of 35?
Let the numbers by 'm' & 'n' Hence m + n = 12 mn = 35 Hence m = 35/n Substitute 35/n + n = 12 Multiply through by 'n' 35 + n^2 = 12n n^2 - 12n + 35 = 0 It is now in Quadratic Form ; Factor ( n - 7)(n - 5) = 0 Hence n= 7 or n= 5 It follows that m = 5 , or m = 7
What is the magnitude of cartesian product?
The Cartesian product of two sets, A and B, where A has m distinct elements and B has n, is the set of m*n ordered pairs. The magnitude is, therefore m*n.
What is the product of 8 and the sum of 6?
mn = 8 ( Product) m + n = 6 (Sum) Hence m = 6 - n Substitute (6 -n)n = 8 Multiply out the brackets 6n - n^(2) = 8 n^(2) - 6n + 8 = 0 It is now in quadratic form and will factor Hence ((n - 2)(n - 4) = 0 n = 2 & n = 4 So '2' & '4' are the two numbers. Verification 2 + 4 = 6 2 x 4 = 8
What are your two numbers if their sum is -5 and their product is 4?
Let the number be 'm' & 'n' Hence Sum = m + n = -5 Product = mn = 4 Algebraically rearrange the product to m = 4/n Substitute into the Sum (4/n) + n = -5 Multiply through by 'n' Hence 4 + n^2 = -5n n^2 + 5n + 4 = 0 It is not in Quadrtatic form . to solve. Hence (n - 1)(n - 4) = 0 n = -1 & n = -4 Are the two numbers.
M plus the product of 6 and n?
m+6n
If the product is 80 and the sum us 21 what are the numbers?
Let the number be 'm' & 'n' Hence Product ; mn = 80 Sum ; m + n = 21 m = 21 - n Substitute (21 - n) n = 80 21n - n^2 = 80 n^2 - 21 n + 80 = 0 It is now in quadratic form to factor. ( n - 16)(n - 5) = 0 Hence the numbers are 5 & 16
Why is the product of two consecutive numbers is always odd?
When multiplied together, an odd number and an even number will always produce an odd number. Two consecutive numbers consist of one odd number and one even number, so their product is always an odd number. N-n-n-no! Any number multiplied by an even number yields an even product. Ever;y even number can be expressed as 2*M for some M. If we multiply 2*M by Y, the product is 2*M*Y = 2*(M*Y) which is even. So the right answer is: They don't! The product of two consecutive numbers is always even. Regards, Bill Drissel
