A)
The four small squares are all 3x3, so the perimeter is 3+3+3+3 = 12.
12 * 4 = 48
We can find the perimeter of the large diamond shape using the Pythagorean theorem, a^2 + b^2 = c^2
a = 4
b = 4
4^2 + 4^2 = c^2
16 + 16 = c^2
32 = c^2
c = 5.65685425
So the perimeter of the large diamond shape is
5.65685425+5.65685425+5.65685425+5.65685425 = 22.627417
48+22.6 = 70.6
B)
The areas of the small squares are 3x3 = 9
9 * 4 = 36
The area of the large diamond shape is
5.65685425 * 5.65685425 = 32
36 + 32 = 68
The answer to your question is 14:9
It landed on red 102 times out of 300 spins
that is 102/300 which can reduce to 17/50 as a fraction
17/50 = 0.34 as a decimal
Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0