<h2>Area of a square = side × side</h2>
Let the side = x

<h3>Length of each side = <u>8</u><u> </u><u>units</u></h3>
.... Hope this will help....
<h3>Given</h3>
tan(x)²·sin(x) = tan(x)²
<h3>Find</h3>
x on the interval [0, 2π)
<h3>Solution</h3>
Subtract the right side and factor. Then make use of the zero-product rule.
... tan(x)²·sin(x) -tan(x)² = 0
... tan(x)²·(sin(x) -1) = 0
This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:
... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)
Then our equation becomes
... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0
... -sin(x)²/(1 +sin(x)) = 0
Now, we know the only solutions are found where sin(x) = 0, at ...
... x ∈ {0, π}
6*28= 168
You put 168 in the green box
it would be covered by tuesday: because if it doubles each day and it is half full on monday it would be double that so it would be 100% 4/4 covered
M=-3.5+2t, R=-3t when they meet M=R so
-3.5+2t=-3t
-3.5+5t=0
5t=3.5
t=0.7 hrs (42 minutes)