There are three kids. the sum of individual squares of their body weights (2 + 2 + 2) is 100. the sum of the product of their weights taking two friends at a time i. e. + + is 150. what will the weighing machine read if all the three kids stand on it at the same time?

20

Step-by-step explanation:

Givens

Let child one = x

Let child two = y

Let child three = z

Equations

x^2 + y^2 + z^2 = 100

xy + xz + yz = 150

Solution

There's a trick here. The square of their weights added together is equal (with some modification) to the given conditions. Start by squaring (x+y+z).

(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz

Take out 2 as a common factor from the last three terms.

(x + y + z)^2 = (x^2 + y^2 + z^2+ 2(xy + xz + yz) )

Substitute the given conditions into the equation. (x^2 + y^2 + z^2) = 100 and 2*(xy + xz + yz) = 2 * 150

(x + y + z)^2 = 100 + 2*150

(x + y + z)^2 = 100 + 300

(x + y + z)^2 = 400

Take the square root of both sides.

sqrt(x+y+z)^2 = sqrt(400)

x + y + z = 20

Note

This answer tells you nothing about the values of x y and z. On the other hand it does not ask for the values of x y and z.

1.3 million hoops i believe

c

step-by-step explanation: