Aweather balloon is launched from the ground at a spot 250 yards from a 65 foot royal palm tree. what is the minimal angle of elevation at which the balloon must take off in order to avoid hitting the tree? assume that the balloon flies in a straight line and the angle of elevation stays constant.

5°

Step-by-step explanation:

Mean of the data set is one of the most commonly used measures of central tendency. Basically, it means the average value of the data set. For finding it, one doesn't have to arrange the data in order. All it takes is to add them all together and divide that by the number of the data. It can be mathematically written as:

(x1 + x2 + x3 + + xn) / n

This obviously means that Ashrita did the best job in calculating the mean of this data set. Let us mark the launching point with A, bottom of the tree with B and top of the tree with C.

In this way, we get a right triangle whose sides are AB = 250 yards (which is 750 feet) and BC = 65 feet.

We want to know the angle CAB (the angle at point A) and we can determine this using trigonometric functions.

Since we know the length of the opposite and the adjacent sides to this angle, we will use tangent of an angle (tan).

tan = opposite side / adjacent side

tan = 65 / 750 = 0.087

Now, knowing the tan of an angle, we can find the angle using the arctan function:

arctan (0.087) = 4.97° which we can round to 5°.

That means that, moving in a straight line and at constant angle, balloon needs to have an angle of elevation of at least 5° in order to avoid hitting the tree.

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!