Given
we have
since we are <em>approaching</em> 
, so effectively 
and we use the definition of 
under that condition.
Answer:
D^2 = (x^2 + y^2) + z^2
and taking derivative of each term with respect to t or time, therefore:
2*D*dD/dt = 2*x*dx/dt + 2*y*dy/dt + 0 (since z is constant)
divide by 2 on both sides,
D*dD/dt = x*dx/dt + y*dy/dt
Need to solve for D at t =0, x (at t = 0) = 10 km, y (at t = 0) = 15 km
at t =0,
D^2 = c^2 + z^2 = (x^2 + y^2) + z^2 = 10^2 + 15^2 + 2^2 = 100 + 225 + 4 = 329
D = sqrt(329)
Therefore solving for dD/dt, which is the distance rate between the car and plane at t = 0
dD/dt = (x*dx/dt + y*dy/dt)/D = (10*190 + 15*60)/sqrt(329) = (1900 + 900)/sqrt(329)
= 2800/sqrt(329) = 154.4 km/hr
154.4 km/hr
Step-by-step explanation:
Answer:
2800 cm^3, or 2.8 liters
Step-by-step explanation:
The base of the tank remains 14 cm × 25 cm. The remaining height is 15 -7 = 8 cm, so the remaining volume is ...
(14 cm)(25 cm)(8 cm) = 2800 cm^3 = 2.8 liters
_____
1 liter = 1000 cm^3
Answer: B.
Using trigonometric ratios, we know that:
tan x = opp/adj
Plug in the values.
tan A = 20/48
The side opposite to angle A is 20, and the adjacent side is 48.
Simplify the fraction.
tan A = 5/12
(5x+1) If X=3 then 5(3)+1=16
(2×2y) if Y=2 then 2×(2)2=8
16÷8=2
