Use Pythagorean theorem:
9i-j = sqrt (9^2 - 1^2) = sqrt(81-1) = sqrt80
now divide both terms in V by that:
u = 9/sqrt(80)i - 1/sqrt(80)j
see attached picture:
The given equations are
(1)
(2)
When t=0, obtain
Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means
.
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means
.
Because
, obtain
Answer:
The slope of the curve at t=0 is 1/6.
Answer: A, C, and E! I just took this on Khan Academy! Please mark me brainliest because it it correct!
Answer:
The time(s) the ball is higher than the building: Interval (0,4)
Step-by-step explanation:
s(t)=-16t^2+64t+400
Determine the time(s) the ball is higher than the building:
s(t)>400
-16t^2+64t+400>400
Subtracting 400 both sides on the inequality:
-16t^2+64t+400-400>400-400
-16t^2+64t>0
Multiplying the inequality by -1:
(-1)(-16t^2+64t>0)
16t^2-64t<0
Fatorizing: Comon factor 16t:
16t(16t^2/16t-64t/16t)<0
16t(t-4)<0
t is greater than zero:
t>0→t-4<0→t-4+4<0+4→t<4
Then t>0 ant t<4:
Solution = (0, Infinite) ∩ (-Infinite, 4)
Solution = (0,4)
