10 and 3
1500 = 30 x 50
I think you mean product... 4 and 9
To find two numbers where the difference is 7 and the product is 30, we can set up the equations: ( x - y = 7 ) (difference) and ( x \cdot y = 30 ) (product). Solving these equations, we can express ( x ) as ( y + 7 ) and substitute it into the product equation to find the values of ( x ) and ( y ). The numbers that satisfy both conditions are 10 and 3, since ( 10 - 3 = 7 ) and ( 10 \cdot 3 = 30 ).
It is 21 and 9. Because you halve 30 to get 15, then add 6 to each to get the difference.
The complex conjugate pair, 4 ± 3.7417i where iis imaginary square root of -1.
1500 = 30 x 50
I think you mean product... 4 and 9
To find two numbers where the difference is 7 and the product is 30, we can set up the equations: ( x - y = 7 ) (difference) and ( x \cdot y = 30 ) (product). Solving these equations, we can express ( x ) as ( y + 7 ) and substitute it into the product equation to find the values of ( x ) and ( y ). The numbers that satisfy both conditions are 10 and 3, since ( 10 - 3 = 7 ) and ( 10 \cdot 3 = 30 ).
It is 21 and 9. Because you halve 30 to get 15, then add 6 to each to get the difference.
Take 13 and 17 as an example. Sum is 13 + 17 ie 30, product is 13 x 17 = 221
Product
How about: 30 and 30
The numbers are 30 and 20.
The complex conjugate pair, 4 ± 3.7417i where iis imaginary square root of -1.
The numbers are 18 and 12
86
10 and 15