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: 28 feet.
Step-by-step explanation:
Given: During the winter months, freshwater fish sense the water getting colder and swim to the bottom of lakes and rivers to find warmer water.
The total depth of the lake = 32 feet
Since, fish swims 7/8 of the depth, then the depth of lake to which the fish swim=
Since, 
therefore, 
Hence, the fish swim 28 feet in lake.
<span>(1,625) No
(0,-25) No
(-1,-1) No
Think about what an integer exponent means for an negative base and you'll understand this problem. For instance the powers of -25 would be
-25^1 = -25
-25^2 = (-25) * (-25) = 625
-25^3 = (-25)*(-25)*(-25) = -15625
and so on, giving 390625, -9765625, 244140625, etc.
But that's a different subject. For the ordered pairs given, let's check them out.
(1,625)
-25^1 + 1 = -25 + 1 = -24. And -24 is not equal to 625, so "No".
(0,-25)
-25^0 + 1 = 1 +1 = 2.
Note: Any real number other than 0 raised to the 0th power is 1. And 2 is not equal to -25, so "No".
(-1,-1)
-25^(-1) + 1 = 1/(-25^1) + 1 = 1/-25 + 1 = 24/25.
And 24/25 is not equal to -1, so also "No".</span>
