Anwser
C
I think
Wait not it’s can’t be c
D. Yeah d jdjddjslkfkd
Answer:
D option
Step-by-step explanation:
Refer to attachment.
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em> </em>
Step-by-step explanation:
draw it through the center to form the simplest line if that helps :)
-2(-2) = 4 ——> (-2, 4)
-2(0) = 0 ——> (0, 0)
-2(2) = -4 ——> (2, -4)
The answer is most likely the first option.
Answer:
The sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is <u>169</u>
Step-by-step explanation:
Given : the difference of the squares of the numbers is 5 and the product of the numbers is 6.
We have to find the sum of the squares of two numbers whose difference and product is given using given identity,
![(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2](https://tex.z-dn.net/?f=%28x%5E2%2By%5E2%29%5E2%3D%28x%5E2-y%5E2%29%5E2%2B%282xy%29%5E2)
Since, given the difference of the squares of the numbers is 5 that is ![(x^2-y^2)^2=5](https://tex.z-dn.net/?f=%28x%5E2-y%5E2%29%5E2%3D5)
And the product of the numbers is 6 that is ![xy=6](https://tex.z-dn.net/?f=xy%3D6)
Using identity, we have,
![(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2](https://tex.z-dn.net/?f=%28x%5E2%2By%5E2%29%5E2%3D%28x%5E2-y%5E2%29%5E2%2B%282xy%29%5E2)
Substitute, we have,
![(x^2+y^2)^2=(5)^2+(2(6))^2](https://tex.z-dn.net/?f=%28x%5E2%2By%5E2%29%5E2%3D%285%29%5E2%2B%282%286%29%29%5E2)
Simplify, we have,
![(x^2+y^2)^2=25+144](https://tex.z-dn.net/?f=%28x%5E2%2By%5E2%29%5E2%3D25%2B144)
![(x^2+y^2)^2=169](https://tex.z-dn.net/?f=%28x%5E2%2By%5E2%29%5E2%3D169)
Thus, the sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169