Answer: 
Step-by-step explanation:
if we have a square with sides 'y' units,
then, its breadth = length = y
If we add 10 to one side it will become (y+10) (longer one) and subtract 10 from another, then it will become (y-10) (shorter one).
Area of rectangle = breadth x length
So, area of rectangle 
![= y^2-10^2=y^2-100\ \ \ \ \ [(a-b)(a+b)=a^2-b^2]](https://tex.z-dn.net/?f=%3D%20y%5E2-10%5E2%3Dy%5E2-100%5C%20%5C%20%5C%20%5C%20%5C%20%5B%28a-b%29%28a%2Bb%29%3Da%5E2-b%5E2%5D)
Hence, the new area as a polynomial =
Multiple the second equation by 2 and subtract it from the first
2x + 5y = 10
2x - 6y = -12
now subtract
11y = 22
y = 2
Plug y = 2 in to either equation
x - 3(2) = -6
x - 6 = -6
x = 0
The point at where they intersect is (0, 2)
Fourth root of 1 is ±1
Hope that helps
Answer:
<QRS = 55 degrees
Step-by-step explanation:
Given the following
m∠QRS=(10x−1) ∘
∠RPQ=(3x+17) ∘
m∠PQR=(2x+12)
External angle m<QRS = 10x-1
Interior angles are m∠RPQ=(3x+17) ∘ and m∠PQR=(2x+12)
Using the law that the sum of interior angle is equal to exterior
10x + 1 = 3x+17 + 2x + 12
10x+1 = 5x+29
10x-5x = 29-1
5x = 28
x = 28/5
x = 5.6
Get m<QRS
Recall that <QRS = 10x - 1
<QRS = 10(5.6) - 1
<QRS = 56-1
<QRS = 55 degrees
A formula is a way of an expression in order to make solving a question easier than ever.
*that's what I think . ...*
