It is 65 over 100. simplified is 13 over 20
The answer is -2, hopes this helps
L = building side
W = non-building side
P = 2W + L = 42 (note only one L because the other L is the building itself)
Solve for L:
L = 42 - 2W
Area = l*w
Area = (42-2W)W = 42W - 2W2
Let area be y, so y = -2W2 +42W
Note this is a parabola pointing down because the coefficient of the W2 is negative. That makes the vertex the maximum for which we are searching.
Vertex of this parabola is at W=-b/2a, if the quadratic is aW2 + bW + c = 0
a = -2
b = 42
W = -42/(2*-2) = -42/-4 = 10.5
W = 10.5
L = 42-2(10.5) = 42-21 = 21
Area = L*W = 21 * 10.5
A = 220.5 ft2
Throughout all of these steps I'm only going to alter the left hand side (LHS). I am NOT going to change the right hand side (RHS) at all.
Before I change the LHS of the original equation, let's focus on the given identity
cot^2(x) + 1 = csc^2(x)
Since we know it's an identity, we can subtract 1 from both sides and the identity would still hold true
cot^2(x) + 1 = csc^2(x)
cot^2(x) + 1-1 = csc^2(x)-1
cot^2(x) + 0 = csc^2(x)-1
cot^2(x) = csc^2(x)-1
So we'll use the identity cot^2(x) = csc^2(x)-1
---------------------------------------------
Now onto the main equation given
cot^2(x) + csc^2(x) = 2csc^2(x) - 1
cot^2(x) + csc^2(x) = 2csc^2(x) - 1 .... note the term in bold
csc^2(x)-1 + csc^2(x) = 2csc^2(x) - 1 .... note the terms in bold
[ csc^2(x) + csc^2(x) ] - 1 = 2csc^2(x) - 1
[ 2csc^2(x) ] - 1 = 2csc^2(x) - 1
2csc^2(x) - 1 = 2csc^2(x) - 1
The bold terms indicate how the replacements occur.
So the original equation has been proven to be an identity because the LHS has been altered to transform into the RHS
4y is 55% of 7.27 i believe?
