Answer:
the taco stand would sell 37 tacos in 1/2 an hour.
They are traveling at right angles to each other so we can say one is traveling north to south and the other west to east. Then we can say that there positions, y and x are:
y=150-600t x=200-800t
By using the Pythagorean Theorem we can find the distance between these two planes as a function of time:
d^2=y^2+x^2, using y and x from above
d^2=(150-600t)^2+(200-800t)^2
d^2=22500-180000t+360000t^2+40000-320000t+640000t^2
d^2=1000000t^2-500000t+62500
d=√(1000000t^2-500000t+6250)
So the rate of change is the derivative of d
dd/dt=(1/2)(2000000t-500000)/√(1000000t^2-500000t+6250)
dd/dt=(1000000t-250000)/√(1000000t^2-500000t+6250)
So the rate depends upon t and is not a constant, so for the instantaneous rate you would plug in a specific value of t...
...
To find how much time the controller has to change the airplanes flight path, we only need to solve for when d=0, or even d^2=0...
1000000t^2-500000t+62500=0
6250(16t^2-8t+1)=0
6250(16^2-4t-4t+1)=0
6250(4t(4t-1)-1(4t-1))=0
6250(4t-1)(4t-1)=0
6250(4t-1)^2=0
4t-1=0
4t=1
t=1/4 hr
Well technically, the controller has t<1/4 because at t=1/4 impact will occur :)
First triangle=bxh/2
=3x4/2
=6in squared
rectangle=l x w
=3 x 10
=30in squared
trapezoid=3+5x2/2
=8in squared
6 + 30 + 8
=44 in squared
Answer:
It is A.
Step-by-step explanation:
To solve for b, use the 45-45-90 triangle theorem, in which each of the legs is x, so the legs would be 8. The hypotenuse would therefore be 8√2.
So without further solving the answer is A, since it's the only one with 8√2.
However, I will still solve for A and C. Using the 30-60-90 theorem, we have the sides as x, x√3, and 2x. The second longest side is b. Using this, we find a = 4√6 and c to be 4√2
