x + y = 28 x*y = 7 x=7/y replacing X in eq 1 7/y+y=28 y^2 - 28y + 7 = 0 using above solutions find the reciprocal and sum -- Dhruv
Not two numbers- the same number twice: 2.5 and 2.5. To find this out, use simultaneous equations: xy=6.25 and x+y=5. Rearrange the second equation to y=5-x and substitute this value for y into the first equation: x(5-x)=6.25. Expand the brackets to get 5x-x2=6.25, subtract 6.25 from both sides to get -x2+5x-6.25=0. Apply the quadratic formula to find the final answer.
the answer is 2500 I am sorry but this answer is wrong. You only added the numbers between 1 and 100. The question is the first 100 odd numbers, you need to add 1 to 197. The correct answer is 10,000.
Writing a program for a sum of sine series requires a rather long formula. That formula is: #include #include #include main() { int i,n,x; .
Given a complex number z = a + bi, the conjugate z* = a - bi, so z + z*= a + bi + a - bi = 2*a. Note that a and b are both real numbers, and i is the imaginary unit: +sqrt(-1).
Any right angled triangle The Theorem state that the "Sum of the square on the hypotenuse (C) is equal to the sum of the squares on the other two sides (A and B)" That is C2=A2+B2
It is 0.5
1/x + 1/y = (y+x)/xy But y + x = sum = 150, and xy = product = 40 So sum of reciprocals = 150/40 = 3.75
No, the product of reciprocals is 1.
Suppose the two numbers are x and y. Then, the sum of THEIR reciprocals is 1/x + 1/y = y/xy + x/xy = (y + x)/xy = 7/25
Suppose the numbers are x and y Then the sum of their reciprocals is 1/x + 1/y = y/xy + x/xy = (y+x)/xy = 10/20 = 1/2
sannie
27
Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
(50 + 15) = 65 (50 x 15) = 750 1/50 + 1/15 = 13/150
Sum of squares? Product?
No - the product of numbers is the answer to a multiplication sum, while the sum of numbers is the answer to an addition sum.
"The sum of a number and three times another number is 18. find the numbers if their product is a maximum?"