You can always draw a right-angled triangle where the sloped side c is the line between the two points of interest, and a and b are the sizes of the horizontal and vertical lines.
Then apply c = √(a²+b²)
6 units Foo
Explanation
Cause it’s just 6 units
Answer: Thx it feels good for people to encourage us.
Step-by-step explanation:
Answer:
NO. Ada is not correct.
Step-by-step explanation:
Using Pythagorean Theorem, find the length of the diagonal of the rectangle and the square, respectively.
✔️Diagonal of the Rectangle:
![a^2 + b^2 = c^2](https://tex.z-dn.net/?f=%20a%5E2%20%2B%20b%5E2%20%3D%20c%5E2%20)
Where,
a = 8 in.
b = 16 in.
c = hypotenuse (longest side of a right ∆)
Plug in the values into the equation
![8^2 + 16^2 = c^2](https://tex.z-dn.net/?f=%208%5E2%20%2B%2016%5E2%20%3D%20c%5E2%20)
![64 + 256 = c^2](https://tex.z-dn.net/?f=%2064%20%2B%20256%20%3D%20c%5E2%20)
![320 = c^2](https://tex.z-dn.net/?f=%20320%20%3D%20c%5E2%20)
Take the square root of both sides
![\sqrt{320} = \sqrt{c^2}](https://tex.z-dn.net/?f=%20%5Csqrt%7B320%7D%20%3D%20%5Csqrt%7Bc%5E2%7D%20)
(nearest tenth)
Length of diagonal SQ = 17.9 in
✔️Diagonal of the Rectangle:
![a^2 + b^2 = c^2](https://tex.z-dn.net/?f=%20a%5E2%20%2B%20b%5E2%20%3D%20c%5E2%20)
Where,
a = 8 in.
b = 8 in.
c = hypotenuse (longest side of a right ∆)
Plug in the values into the equation
![8^2 + 8^2 = c^2](https://tex.z-dn.net/?f=%208%5E2%20%2B%208%5E2%20%3D%20c%5E2%20)
![64 + 64 = c^2](https://tex.z-dn.net/?f=%2064%20%2B%2064%20%3D%20c%5E2%20)
![128 = c^2](https://tex.z-dn.net/?f=%20128%20%3D%20c%5E2%20)
Take the square root of both sides
![\sqrt{128} = \sqrt{c^2}](https://tex.z-dn.net/?f=%20%5Csqrt%7B128%7D%20%3D%20%5Csqrt%7Bc%5E2%7D%20)
(nearest tenth)
Length of diagonal OM = 11.3 in.
SQ is not two times the length of OM.
Therefore, Ada is not correct.
For thsi you need to think of 2 simulatneous equations=
16f + 28s= 24232
f + s = 1090
We need to find a way to subtract one equation from the other so you can multiply the bottom by 16
so you would get
16f + 28s = 24232
16f + 16s = 17440
Now you can subtract the second one from the bottom one to get
12s= 6792
s= 566
So there were 566 tickets sold on staurday so to find fridasy you can do 1090 -566, which is 524
so on friday night 524 were sold and on saturday there was 566 sold