Answer:
decreasing at 390 miles per hour
Step-by-step explanation:
Airplane A's distance in miles to the airport can be written as ...
a = 30 -250t . . . . . where t is in hours
Likewise, airplane B's distance to the airport can be written as ...
b = 40 -300t
The distance (d) between the airplanes can be found using the Pythagorean theorem:
d^2 = a^2 + b^2
Differentiating with respect to time, we have ...
2d·d' = 2a·a' +2b·b'
d' = (a·a' +b·b')/d
__
To find a numerical value of this, we need to find the values of its variables at t=0.
a = 30 -250·0 = 30
a' = -250
b = 40 -300·0 = 40
b' = -300
d = √(a²+b²) = √(900+1600) = 50
Then ...
d' = (30(-250) +40(-300))/50 = -19500/50 = -390
The distance between the airplanes is decreasing at 390 miles per hour.
75, 76, 80, 83, 84, 85, 89, 91, 94, 94
The median is 84.5
You add the two middle numbers (84, 85) and divide by 2.
<h3>
Answer:</h3>
See the attached
<h3>
Step-by-step explanation:</h3>
When you square the binomial (a -b), you get ...
... (a -b)² = a² -2ab +b²
That is, both the a² and b² terms have positive signs, and the middle term is twice the product of the roots of the squared terms.
The last two selections have negative signs on the constant, so cannot be perfect square trinomials.
The first selection has a middle term that is -ab, not -2ab, so it is not a perfect square trinomial, either.
The second selection is the correct one:
... 4a² -20a +25 = (2a +5)²
Answer:
0.4
Step-by-step explanation:

<span>First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.
You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in </span><span>signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...
The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so </span>

<span> where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.</span>