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:
The difference of -34 and -43 is 9
-34 - -43 = 9
I hope this helps
Answer: 16875
Step-by-step explanation:
Let the price of the washing machine be x.
Since she loses 20% in the bargain, this implies that she sold it at (100% - 20%) = 80%.
Therefore, 80% of x = 13500
0.8 × x = 13500
0.8x = 13500
Divide both side by 0.8
0.8x/0.8 = 13500/0.8
x = 16875
Therefore, the price at which she bought the washing machine is 16875 rupees.
I don’t know if this will help....
It is c I think I'm not positive but Ive done this one
