Answer:
d=16t squared
Step-by-step explanation:
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
A = 16 over 49
b= 2 then 1 over 10
Pemdas
parenthasees exponents mult/division additon/subtraction
parethenasees
x-3 and x+5 we can't do anything with so next
5(x-3)=5x-15
2(x+5)=2x+10
5x-15+2=5x-13
(2x+10-9)=2x+1
(5x-13)-3(2x+1)
-3(2x+1)=-6x-3
5x-13-6x-3=-x-16
4(-x-16)=-4x-64
the answer is -4x-64