Answer: If we define 2:00pm as our 0 in time; then:
at t= 0. the velocity is 30 mi/h.
then at t = 10m (or 1/6 hours) the velocity is 50mi/h
Then, if we think in the "mean acceleration" as the slope between the two velocities, we can find the slope as:
a= (y2 - y1)/(x2 - x1) = (50 mi/h - 30 mi/h)/(1/6h - 0h) = 20*6mi/(h*h) = 120mi/
Now, this is the slope of the mean acceleration between t= 0h and t = 1/6h, then we can use the mean value theorem; who says that if F is a differentiable function on the interval (a,b), then exist at least one point c between a and b where F'(c) = (F(b) - F(a))/(b - a)
So if v is differentiable, then there is a time T between 0h and 1/6h where v(T) = 120mi/
Answer:
<h2>y = 2</h2>
Step-by-step explanation:
Answer:
y = -34x + 62
Step-by-step explanation:
<u>Use the point slope form: (y - y1) = m(x - x1)</u>
<u />
(y - (-6) = -34(x - 2)
y + 6 - 6 = -34x + 68 - 6
y = -34x + 62
Answer: y = -34x + 62
Answer:
(2, - 3 )
Step-by-step explanation:
Given the 2 equations
2x - 3y = 13 → (1)
x + 2y = - 4 → (2)
Rearrange (2) expressing x in terms of y by subtracting 2y from both sides
x = - 4 - 2y → (3)
Substitute x = - 4 - 2y into (1)
2(- 4 - 2y) - 3y = 13 ← distribute and simplify left side
- 8 - 4y - 3y = 13
- 8 - 7y = 13 ( add 8 to both sides )
- 7y = 21 ( divide both sides by - 7 )
y = - 3
Substitute y = - 3 into (3) for corresponding value of x
x = - 4 - 2(- 3) = - 4 + 6 = 2
Solution is (2, - 3 )
The answer is B)422 ,you're welcome :)
