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:
u= 230
Step-by-step explanation:
Answer:
This question is a rather easy one which can be answered by simply solving the embedded quadratic equation.
f(x) = -
+ 6x + 16
To make our solving easier, what we do is multiply the whole equation by -1. By so doing, we make "x" positive. Then, we have
f(x) = - 6x - 16
Solving this equation quadratically, we have
+ 2x - 8x - 16 = and when we factorize this expression, we have
x(x + 2) -8(x + 2) =
(x - 8) (x + 2) = 0
x - 8 = 0, x + 2 = 0
x = 8, or x = -2
Since distance can not be in negative(minus), we then have our x to be 8 meters.
Another example on factorization can be seen here brainly.com/question/13813750
A+b+c=130
a,b,c ip cu 2,3,4
2a=3b=4c=k
a=k/2
b=k/3
c=k/4
k/2 + k/3+ k/4= 130 aducem la acelasi numitor (12)
6k+4k+3k=1560
13k=1560
k=120
a=120/2=60
b=120/3=40
c=120/4=30
