Answer:
there is no change
and hay it's me again. hope this helps
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.
Louis = x
Jane = 2x (Twice as many as Louis)
Together they have 360
x + 2x = 360
3x= 360
x=120
2(120)= 240
Jane addresses 240 envelopes
Answer:
Approximately 55ft
Step-by-step explanation:
Answer:
a) F
b) B, E, D
Step-by-step explanation:
a) The segment with the greatest gradient has the largest change in y-values per unit change in x-values
From the given option, the rate of change of the <em>y </em>to the<em> </em>x-values of B = the gradient = (4 units)/(2 units) = 2
The gradient of F = (-3units)/(1 unit) = -3
The gradient of A = 4/4 = 1
The gradient of C = -2/5
The gradient of D = 2/6 = 1/3
The gradient of E = 3/4
The segment with the greatest gradient is F
b) The steepest segment has the higher gradient
From their calculated we have;
The gradient of segment B = 2 therefore, B is steeper than E that has a gradient of 3/4, and E is steeper than D, as the gradient of D = 1/3
Therefore, we have;
B, E, D.
