Solve by back-calculating for the missing number . . . (58 - 7 - 14 + 3)/(5 × 2) = 4
The 18th sequence is - 41
Answer:
Step-by-step explanation:
Since this problem talks about rates of change, then the concept of calculus is very useful. But first, let's find at least two equations in order to solve this system. The first one is the area of a triangle written as
A = 1/2 ab sin θ, where a and b are the sides that from the angle θ. So, we substitute a=6 and b=10. That makes it:
A = 1/2 (6)(10)sin θ = 30 sin θ
Now, you differentiate implicitly (both sides simultaneously) with respect to time.
dA/dt = 30 cosθ (dθ/dt)
We replace dθ/dt = 0.06 rad/s, as mentioned in the problem. Then, the rate of change of the area of the triangle when θ = π/3 rad with respect to time is
dA/dt = 30cos(π/3) (0.06)
dA/dt = 1.8 m²/s
Therefore, the rate of change of the area of the triangle is 1.8 m² per second.
Answer:
80 feet
Step-by-step explanation:
Given:
Initial speed of the car (
) = 40 ft/sec
Deceleration of the car (
) = -10 ft/sec²
Final speed of the car (
) = 0 ft/sec
Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.
Now, deceleration means rate of decrease of velocity.
So, 
Negative sign means the velocity is decreasing with time.
Now,
using chain rule of differentiation. Therefore,

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,
![\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_%7B40%7D%20%7Bv%7D%20%5C%2C%20dv%3D%5Cint%5Climits%5Ex_0%20%7B-10%7D%20%5C%2C%20dx%5C%5C%5C%5C%5Cleft%20%5B%20%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5Cright%20%5D_%7B40%7D%5E%7B0%7D%3D-10x%5C%5C%5C%5C-10x%3D%5Cfrac%7B0%7D%7B2%7D-%5Cfrac%7B1600%7D%7B2%7D%5C%5C%5C%5C10x%3D800%5C%5C%5C%5Cx%3D%5Cfrac%7B800%7D%7B10%7D%3D80%5C%20ft)
Therefore, the car travels a distance of 80 feet before stopping.