Answer:
10 is your answer
Step-by-step explanation:
Note that the side given to us has a measurement of 5 for one of the legs of the isosceles.
This means that the <em>1</em> side on the 30-60-90 triangle has a measurement of 5.
Now, note the measurements of the triangle sides for a 30-60-90 triangle. They measure at: 1 , √3 , 2.
the side 1 is given to us, at 5. You are solving for the hypothenuse (2). Multiply 2 to the side 1
5 * 2 = 10
10 is your answer
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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
Sq32/8 = sq4
It would be c. 2
Answer:
48
Step-by-step explanation:
used a calculator
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