What is nbr2 at the end of the program segment execution? nbr2 = 0 calc = 1 x = 7 while x >=3 calc = calc * x x = x - 2
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Answer:
Shortest distance from A to C = 102.9005 m.
Step-by-step explanation:
It is given that, ABCD is a rectangular park.
The length of the park is 80 m.
The breadth of the park is 50 m.
The diameter of the circle = 15 m.
We have to calculate the shortest distance from A to C across the park.
The distance AC = = 94.339 m.
As one has to pass only through the lines shown, he cannot pass through the circle.
So, we have to subtract the diameter of 15 m from AC
=> 94.339 m - 15 m = 79.339 m.
One must pass through either half of the circumference of the circle.
Since, diameter of circle = 15 m, its radius(r) = = 7.5 m.
The circumference of the circle = 2×π×r = 47.123 m.
Half of the circumference = = 23.5615 m.
Distance from A to C passing through circumference = 79.339 m + 23.5615 m = 102.9005 m.
As we have to calculate the shortest distance from A to C, one cannot pass from A to C either through B or D,
since the distance ABC or ADC = 50+80 = 130 m.
Therefore, shortest distance from A to C = 102.9005 m.
An obtuse triangle has one obtuse angle, or an angle that measures more than 90° but less than 180°. Options 1 and 2 are not correct, because these are acute angles.
So, one of the angles must be 40°, and the other may be 100° or 120°.
The measures of the angles in a triangle add up to 180°.
The other two angles are 40° and 100°.
So, one of the angles must be 40°, and the other may be 100° or 120°.
The measures of the angles in a triangle add up to 180°.
The other two angles are 40° and 100°.