<span>2n^2 - 7n - 3 = 0
a = 2
b = -7
c =-3
Then use the Quadratic formula:
x = [-b +-sqroot(b^2 -4*a*c)] / 2*a
</span>
dividido por la epotenusa es 7 :v
It’s graph 2 because it is. There is a pause for the taking the bath and then the emptiest mhm so it can’t be graph 3
To solve this system of equations, since y is already isolated in both equations, you can set the expressions to equal each other to solve for x:
-x + 2 = - 5x - 6
-x + 5x = -6 - 2
4x = -8
x = -2
Now that we have x, we can substitute it into one of the equations to find y:
y = -(-2) + 2
y = 2 + 2
y = 4
The last step is to substitute both values into both equations to see if they are correct:
4 = -(-2) + 2 --> 4 = 2 + 2 <--True
4 = -5(-2) - 6 --> 4 = 10 - 6 <--True
Answer:
x = -2
y = 4
Answer: 2 meters.
Step-by-step explanation:
Let w = width of the cement path.
Dimensions of pool : Length = 15 meters , width = 9 meters
Area of pool = length x width = 15 x 9 = 135 square meters
Along width cement path, the length of region = 
width = 
Area of road with pool = 
Area of road = (Area of road with pool ) -(area of pool)
![\Rightarrow\ 112 =4w^2+48w+135- 135\\\\\Rightarrow\ 112= 4w^2+48w\\\\\Rightarrow\ 4 w^2+48w-112=0\\\\\Rightarrow\ w^2+12w-28=0\ \ \ [\text{Divide both sides by 4}]\\\\\Rightarrow\ w^2+14w-2w-28=0\\\\\Rightarrow\ w(w+14)-2(w+14)=0\\\\\Rightarrow\ (w+14)(w-2)=0\\\\\Rightarrow\ w=-14\ or \ w=2](https://tex.z-dn.net/?f=%5CRightarrow%5C%20112%20%3D4w%5E2%2B48w%2B135-%20135%5C%5C%5C%5C%5CRightarrow%5C%20112%3D%204w%5E2%2B48w%5C%5C%5C%5C%5CRightarrow%5C%204%20w%5E2%2B48w-112%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B12w-28%3D0%5C%20%5C%20%5C%20%5B%5Ctext%7BDivide%20both%20sides%20by%204%7D%5D%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B14w-2w-28%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%28w%2B14%29-2%28w%2B14%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%28w%2B14%29%28w-2%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%20w%3D-14%5C%20or%20%5C%20w%3D2)
width cannot be negative, so w=2 meters
Hence, the width of the road = 2 meters.
