Answer:
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.
I’m pretty sure it is 1/36!
The Lea thing of ce are 1.25
We are required to find the price of the first game
The price of the first game is $20
Given:
let
price of the first game = x
price of the second game = x + 50% of x
= x + 0.5x
= 1.5x
Price of the third game = 50% of 1.5x
= 0.5 × 1.5x
= 0.75x
Total cost of game = $65
<em>Total cost of game = price of the first game + price of the second game + Price of the third game</em>
65 = x + 1.5x + 0.75x
65 = 3.25x
divide both sides by 3.25
x = 65 ÷ 3.25
x = 20
Therefore, the price of the first game is $20
Read more:
brainly.com/question/291220
