8x^6 + 4x^2 -x+213
The 3 and the 210 would be added together
F(-1) = -(-1) - 2 f(2) = -(2) - 2 f(4) = -(4) - 2
f(-1) = 1 - 2 f(2) = -2 - 2 f(4) = -4 - 2
f(-1) = -1 f(2) = -4 f(4) = -6
Answer:
-390 mph
Step-by-step explanation:
Let a and b represent, respectively, the distances of A and B from the airport. The distance d between the planes is then given by the Pythagorean theorem as ...
d² = a² + b²
Differentiating with respect to time, we have ...
2d·d' = 2a·a' +2b·b'
Solving for d', we get ...
d' = (a/d)a' +(b/d)b'
The value of d at the time of interest is ...
d = √(a² +b²) = √(30² +40²) = √2500 = 50
Then the rate of change of separation is ...
d' = (30/50)(-250 mph) +(40/50)(-300 mph) = (-150 -240) mph
d' = -390 mph
The distance between planes is decreasing at 390 miles per hour.
Answer:
N+P or number 3
Step-by-step explanation:
$4,000 in the 10% per year account
$11,000 in the 12% per year account
