Some people advise that in very cold weather, you should keep the gas tank in your car more than half full. Irene's car had 5.
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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
Answer: The calculations are done below.
Step-by-step explanation:
(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,
Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.
(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,
Since, BC = CA, so the triangle ABC will be an isosceles triangle.
(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,
Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.
(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,
Since AB² + CA² = BC², so this will be a right angled triangle.
(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,
Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.
Thus, the match is done.