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.
Your question has been heard loud and clear.
2^0= 1
5^1= 5
4^3=64
2^0+5^1+4^3/7= 1+5+64/7 = 6+64/7= 15.14
So , 2^0+5^1+4^3/7 = 15.14
Thank you
The answer to your question is 3
Answer:
D, (1,2)
Step-by-step explanation:
It all comes down to substitution. In this case the coefficient of x is 3 and the coefficient of y is 4. The format of these coordinates being (x,y).
1. Plug in your x value (1 in this circumstance) and solve:
3(1) + 4y < 12
3 + 4y < 12
2. Plug in your y value (2 in this circumstance) and solve:
3 + 4(2) < 12
3 + 8 < 12
3. Solve
3 + 8 = 11
11 < 12
You have to take the 2 divided by 12. And 12 divided by 2 is 6. So the answer is 6
