$5.49 + $10.49=$15:49 there’s your answer
Answer:
The average velocity of the ball at the given time interval is -122.3 ft/s
Step-by-step explanation:
Given;
velocity of the ball, v = 34 ft/s
height of the ball, y = 34t - 26t²
initial time, t₀ = 3 seconds
final time, t = 3 + 0.01 = 3.01 seconds
At t = 3 s
y(3) = 34(3) - 26(3)² = -132
The average velocity of the ball in ft/s is given as;
![V_{avg} = \frac{y(3.01)-y(3)}{3.01 -3}\\\\V_{avg} = \frac{34(3.01)-26(3.01)^2-y(3)}{3.01 -3}\\\\V_{avg} = \frac{-133.223-y(3)}{0.01}\\\\V_{avg} = \frac{-133.223-(-132)}{0.01}\\\\V_{avg} =\frac{-1.223}{0.01}\\\\V_{avg} = -122.3 \ ft/s](https://tex.z-dn.net/?f=V_%7Bavg%7D%20%3D%20%5Cfrac%7By%283.01%29-y%283%29%7D%7B3.01%20-3%7D%5C%5C%5C%5CV_%7Bavg%7D%20%3D%20%5Cfrac%7B34%283.01%29-26%283.01%29%5E2-y%283%29%7D%7B3.01%20-3%7D%5C%5C%5C%5CV_%7Bavg%7D%20%3D%20%5Cfrac%7B-133.223-y%283%29%7D%7B0.01%7D%5C%5C%5C%5CV_%7Bavg%7D%20%3D%20%5Cfrac%7B-133.223-%28-132%29%7D%7B0.01%7D%5C%5C%5C%5CV_%7Bavg%7D%20%3D%5Cfrac%7B-1.223%7D%7B0.01%7D%5C%5C%5C%5CV_%7Bavg%7D%20%3D%20-122.3%20%5C%20ft%2Fs)
Therefore, the average velocity of the ball at the given time interval is -122.3 ft/s
Answer:
CD ≈ 15.0
Step-by-step explanation:
calculate CD using the distance formula
d = ![\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%29%5E2%2B%28y_%7B2%7D-y_%7B1%7D%29%5E2%20%20%20%20%7D)
with (x₁, y₁ ) = C (7, - 4) and (x₂, y₂ ) = D (- 8, - 5)
CD = ![\sqrt{(-8-7)^2+(-5-(-4))^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-8-7%29%5E2%2B%28-5-%28-4%29%29%5E2%7D)
= ![\sqrt{-15)^2+(-5+4)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B-15%29%5E2%2B%28-5%2B4%29%5E2%7D)
= ![\sqrt{225+(-1)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B225%2B%28-1%29%5E2%7D)
= ![\sqrt{225+1}](https://tex.z-dn.net/?f=%5Csqrt%7B225%2B1%7D)
= ![\sqrt{226}](https://tex.z-dn.net/?f=%5Csqrt%7B226%7D)
≈ 15.0 ( to the nearest tenth )
Conversion!
1 kg = 1000 g.
3 kg = 3000 g.
9 kg = 9000 g.
9000 - 3000 = 6000 g.
Angie's puppy gained 6000 g (6 kg.)
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