Can anyone help me wit this!!
1 answer:
Answer:
- See attachment for table values
- y₁ = y₂ for x = 6
Step-by-step explanation:
In each case, put the x-value in the formula and do the arithmetic. If you're allowed, you can save some time and effort by realizing that the solution (x) will have to be an even number.
y₁ is an integer value for all integer values of x. y₂ is an integer value for even values of x only. y₁ and y₂ will both be integers (and possibly equal) only when x is even.
For example, for x = 6, we have
... y₁ = 3·6 - 8 = 18 -8 = 10
... y₂ = 0.5·6 +7 = 3 +7 = 10
That is, for x = 6, both columns of the table have the same number (10). That is, y₁ = y₂ for x = 6. The solution to the equation
... y₁ = y₂
is
... x = 6.
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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.