When you find the mean you add all of them together. Then you divide your answer by how many numbers there are that you added.
So what I did was add 29+30+28+32+29+31+27= 205
then divide 205 by 7 and you get 29.285
X² + y² - 8x - 12y + 52 = 36
x² - 8x + y² - 12y + 52 = 36
x² - 8x + y² - 12y = 88
(x² - 8x + 16) + (y² - 12y + 36) = 88 + 16 + 36
(x - 4)² + (y - 6)² = 138
(h, k) = (x, y) = (4, 6)
X=2
its x=2 because you have to divide -20x by -40 so the two negatives cancel out each other and then you get 2.
Answer:
rate of the plane in still air is 33 miles per hour and the rate of the wind is 11 miles per hour
Step-by-step explanation:
We will make a table of the trip there and back using the formula distance = rate x time
d = r x t
there
back
The distance there and back is 264 miles, so we can split that in half and put each half under d:
d = r x t
there 132
back 132
It tells us that the trip there is with the wind and the trip back is against the wind:
d = r x t
there 132 = (r + w)
back 132 = (r - w)
Finally, the trip there took 3 hours and the trip back took 6:
d = r * t
there 132 = (r + w) * 3
back 132 = (r - w) * 6
There's the table. Using the distance formula we have 2 equations that result from that info:
132 = 3(r + w) and 132 = 6(r - w)
We are looking to solve for both r and w. We have 2 equations with 2 unknowns, so we will solve the first equation for r, sub that value for r into the second equation to solve for w:
132 = 3r + 3w and
132 - 3w = 3r so
44 - w = r. Subbing that into the second equation:
132 = 6(44 - w) - 6w and
132 = 264 - 6w - 6w and
-132 = -12w so
w = 11
Subbing w in to solve for r:
132 = 3r + 3(11) and
132 = 3r + 33 so
99 = 3r and
r = 33
