About 4.1 seconds. How long was the ball in the air? We are told that t represents time in seconds since the ball was thrown, so it started to be 'in the air' at t = 0 To answer the question, then, we need to know the time when it stopped being in the air. We are told that the ball hit the ground. So that's what happened when it stopped being airborne. We need to relate that event to the mathematics we're working with. What can we say about h , the height of the ball when the ball hits the ground? Answer: The height will be 0 when the ball stops being in the air. Now translate this back to the mathematics: The ball is in the air from t = 0 until the time t when h = 0 . Find the time t that makes h = 0 . That means: solve: − 5 t 2 + 20 t + 2 = 0 We can solve this by solving: 5 t 2 − 20 t − 2 = 0 (Either multiply both sides of the equation by − 1 , or add 5 x 2 − 20 x and − 2 to both sides and then re-write it the other way around) That's a quadratic equation, so try to factor first. But don't spend too much time trying to factor, because not every quadratic is easily factorable and that's OK, because we still have the quadratic formula if we need it. We do need it. t = − ( − 20 ) ± √ ( − 20 ) 2 − 4 ( 5 ) ( − 2 ) 2 ( 5 ) = 20 ± √ 440 10 = 20 ± √ 4 ( 110 ) 10 = 20 ± 2 √ 110 10 = 2 ( 10 ± √ 110 ) 2 ( 5 ) = 10 ± √ 110 5 We can see that 10 < √ 110 < 11 . In fact ( 10 + 1 2 ) 2 = 10 2 + 10 + 1 4 = 110.25 Using 10.25 as an approximation for √ 110 , we get : for the solution t = 10 − √ 110 5 we'll get a negative t . That doesn't make sense. The other solution gives t ≈ 10 + 10.25 5 = 20.5 5 = 4.1 seconds. So the ball was in the air from t = 0 until about t = 4.1 . The elapsed time is the difference, 4.1 seconds.
Answer:
a)
b)
Step-by-step explanation:
From the question we are told that
The Function
Generally the differentiation of function f(x) is mathematically solved as
Therefore
Generally critical point is given as
Generally the maximum and minimum x value for critical point is mathematically solved as
Where
Maximum value of x
Minimum value of x
Therefore interval of increase is mathematically given by
Therefore interval of decrease is mathematically given by
Generally the second differentiation of function f(x) is mathematically solved as
Generally the point of inflection is mathematically solved as
Therefore inflection points is given as
a)Generally the concave upward interval X is mathematically given as
b)Generally the concave downward interval Y is mathematically given as
Answer:
198.6 with the remaining 6
Step-by-step explanation:
596/3
596/3 = 198.666666667 as a decimal form
596/3 = 198.67 in 2 decimal places
596/3 = 198.7 to the nearest tenth
596/3 = 198.67 to the nearest hundredth
596/3 = 198.667 to the nearest thousandth
First you need to find the common denominator. both 6 and 7 go in to 42 evenly
multiply 5/6 by 7 to get 35/42
multiply 2/7 by 6 to get 12/42
subtract 12 from 35 to get 23
23/42
