Answer:
x = 3
Step-by-step explanation:
Mathematically, the mean is the sum of the numbers divided by the count of the numbers
The count of the numbers is 5
Thus;
( 4 + 8 + 9 + x + 2x)/5 = 6
(21 + 3x)/5 = 6
21 + 3x = 6(5)
21 + 3x = 30
3x = 30-21
3x = 9
x = 9/3
x = 3
so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant
![\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx%7D%7B9%7D~~%2C~~%5Cstackrel%7By%7D%7B-3%7D%29%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B9%5E2%2B%28-3%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B81%2B9%7D%5Cimplies%20c%3D%5Csqrt%7B90%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
x=1
Step-by-step explanation:
By factoring, you'll compress the equation to (x-1)^2=0
Next take out the square: x-1=0
Now subtract one to the other side to get x! x=1
Hope this helps! :)
To solve this problem you must apply the proccedure shown below:
1. You have that the ratio of the tangent function is:
tan(angle)=opposite/adjacent
2. Keeping this on mind, you have:
tan(A)=12/5
Therefore, the answer is:
tan(A)=12/5
Answer:
- plane: 550 mph
- wind: 50 mph
Step-by-step explanation:
If p and w represent the speeds of the plane and wind, respectively, the speed into the wind is ...
p - w = (3000 mi)/(6 h) = 500 mi/h
And, the speed with the wind is ...
p + w = (3000 mi)/(5 h) = 600 mi/h
Adding these two equations gives us ...
2p = 1100 mi/h
p = 550 mi/h . . . . . . . divide by 2
Then the wind speed is ...
w = 600 mi/h - p = (600 -550) mi/h
w = 50 mi/h
The rate of the plane in still air is 550 mi/h; the rate of the wind is 50 mi/h.
