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:
5a^2 - 6a -4 - (<span>-7a^2 + 3a -9)
=</span>5a^2 - 6a -4 + 7a^2 - 3a + 9
=12a^2 - 9a + 5
answer
12a^2 - 9a + 5
Answer:
coefficient can be taken of a particular number or variable not a whole expression
Step-by-step explanation:
this may help you
coefficient of x is 3
hope it helps you
please mark me as brainlist
No because (-7, 3) is like (x, y)
Plug and chug [what I call it] into the equation
(3) - 4(-7) < 5
3- (-28)<5
31<5 <—— this is not true so it would be galse
Answer:
The correct answer is = p = 15 and q = 6.
Step-by-step explanation:
Given:
Perimeter of rectangle = 42 cm
p - q = 9 cm
length = p
width = q
we know:
Perimeter of rectangle= 2(l+w)
solution:
42 = 2(P+q)
21= p+q
Difference of p and q = 9 cm.
Then, P+q=21 .... 1
P-q=9 ....2
Adding both 1 and 2
2P = 30
P= 15cm and
q = 15-9
= 6cm
