Answer:
The curvature is
The tangential component of acceleration is
The normal component of acceleration is
Step-by-step explanation:
To find the curvature of the path we are going to use this formula:
where
is the unit tangent vector.
is the speed of the object
We need to find , we know that so
Next , we find the magnitude of derivative of the position vector
The unit tangent vector is defined by
We need to find the derivative of unit tangent vector
And the magnitude of the derivative of unit tangent vector is
The curvature is
The tangential component of acceleration is given by the formula
We know that and
so
The normal component of acceleration is given by the formula
We know that and so
Answer:
350 students
Step-by-step explanation:
Let:
a = number of adults attending
s = number of students attending
a + s = 950 (1) (given)
9a + 6.5s = 7675 (2) (cost per adult * number of adults + cost per student * number of students = total amount spent)
Looking for s. Multiply (1) by 9 then subtract eq (2), this will eliminate the variable a:
(1) 9a + 9s = 8550
- (2) 9a + 6.5s = 7675
-------------------------------
0a + 2.5s = 875
s = 875/2.5 = 350 students
Answer:
Step-by-step explanation:
take derivative with respect to t
derivative of e^t is e^t
derivative of -1.4t is -1.4
Apply chain rule
To figure this out plug in 4 for x
f(4)=5/3(4)^3/2+8(4)^1/2 = 5/3(8) + 8(2)
= 5/3 x 8 + 16 = 40/3 +16 = 88/3
Hope this helps!
We need to follow the order of operations to get this one right.
16 + (10/5) - (3*2)
16 + (2) - (6)
16 + 2 - 6
18 - 6
12<em /><em>. </em>Your answer is true!