IM DESPERATE I need help with #3. I would like it if you could explain how to do it because there are 2 more that are similar to
#3. Thank you!
2 answers:
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Answer:
Our equation for the height is:
y(t) = 275 - 16*t^2.
a) To find the average velocity between two times, t1 and t2, (where t2 > t1) the equation is:
Then:
i) t1 = 4s, t2 = 4s + 0.1s = 4.1s
The average velocity is:
And the units will be ft/s, so the average speed is:
-129.6 ft/s
The minus sign is because te pebble is falling down.
ii) t1 = 4s, t2 = 4s + 0.05s = 4.05s
The average velocity is:
So the average speed is -128.9 ft/s
iii) t1 = 4s, t2 = 4s + 0.01s = 4.01s
The average speed is:
The average speed is -128.16 ft/s.
b) The instantaneous velocity of the pebble after 4 seconds can be obtained by looking at the velocity equation, that is the derivative of the height equation.
v(t) = dy(t)/dt.
v(t) = -2*16*t + 0
Then the velocity at t = 4s is:
v(4s) = -32*4 = -128
The instantaneous velocity at t = 4s is -128 ft/s.
Answer:
part a and b only i was actually looking for c
Step-by-step explanation:
for part b
Area of DEF= 1/2× base × height = 1/2× 2 × 1 = 1
Area of FGD= 1/2× base × height = 1/2× 2.24 × 1.34 =1.50
Area of GBD= 1/2× base × height = 1/2× 2 × 1 = 1
Area of ABG= 1/2× base × height = 1/2× 4.12 × 0.49 =1.01
Area of BCD= 1/2× base × height = 1/2× 1.41 × 2.12 =1.49
The approximate area of the original polygon is 6 square units
