a - length of side of a square
t - length of side of a triangle
The perimeter of a square: 
The perimeter of a triangle: 
We have the area of a triangle: 
The formula of an area of an equilateral trinagle: 
Substitute:
<em>multiply both sides by 4</em>
<em>divide both sides by
</em>

The perimeter of a triangle: 
Substitute to the formula of a perimeter of a square:
<em>divide both sides by 4</em>

The formula of a diagonal of a square: 
Substitute:

Ypu would subtract Y from the other points Y

Then you would subtract X from the ither points X.

Then you would put the Y value over the X

Then simiplify
Answer: Then m = 10<span>°</span>
The cost of a senior citizen ticket is $15 and the cost of a student ticket is $12.
How did I get this?
We know that 6 citizen tickets and 7 student stickers sold for $174 the first day. And 10 citizen tickets and 14 student tickets sold for $318 the second day.
1. create two equations out of this: C= citizen cost per ticket and S = student cost per ticket.
6C + 7S = $174
10C + 14S = $318
2. Use process of elimination. Multiply the first equation by 2 because we want two variables to cancel out.
-12C - 14S = -$348
10C + 14S = $318
Combine like terms.
-2C = $30
Divide by -2 on both sides. The left side cancels out.
C = $30/-2
C = -$15 (In this case the negative doesn't matter)
C = $15 (cost of senior citizen ticket)
Plug the value of C into any of the two equations so we can get the value of S.
6($15) + 7S = $174
Distribute the 6 into the parenthesis.
$90 + 7S = $174
Subtract both sides by $90 and the left side will cancel out.
7S = $84
Divide both sides by 7.
S = $12
Student ticket: $12
Senior citizen ticket: $15
Answer:
The short answer is there isn’t.
Start by writing each of these as an expression:
x * y = 60
x + y = 7
Next, solve each for the same variable; in this case, y:
(x * y) / x = 60 / x
.: y = 60 / x
(x + y) - x = 7 - x
.: y = 7 - x
Next, replace y of the second expression to the first
y = 60 / x & y = 7 - x
.: 7 - x = 60 / x
Now, solve for x:
(7 - x) * x = (60 / x) * x
.: x * 7 - x^2 = 60
This is quadratic, so write it in the form of ax2 + bx + x = 0
(-1)x^2 + (7)x + (-60) = 0
.: a = -1, b = 7, c = -60
Finally solve for b:
x = (-b +- sqrt(b^2 - 4*a*c)) / 2a
.: x = (-7 +- sqrt(7^2 - 4*-1*-60)) / (2 * -1)
.: x = (-7 +- sqrt(49 - 240)) / -2
.: x = (-7 +- sqrt(-191)) / -2
The square root of a negative value is imaginary and thus there’s no real answer to this problem.