Answer:
C = (2,2)
Step-by-step explanation:
B = (10 ; 2)
M = (6 ; 2)
C = (x ; y )
|___________|___________|
B (10;2) M (6;2) C ( x; y)
So:
dBM = dMC
√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]
(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2
0 + (-4)^2 = (y-2)^2 + (x - 6)^2
16 = (y-2)^2 + (x - 6)^2
16 - (x - 6)^2 = (y-2)^2
Also:
2*dBM = dBC
2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]
4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2
4*(16) = (y-2)^2 + (x - 10)^2
64 = (y-2)^2 + (x - 10)^2
64 = 16 - (x - 6)^2 + (x - 10)^2
48 = (x - 10)^2 - (x - 6)^2
48 = x^2 - 20*x + 100 - x^2 + 12*x - 36
48 = - 20*x + 100 + 12*x - 36
8*x = 16
x = 2
Thus:
16 - (x - 6)^2 = (y-2)^2
16 - (2 - 6)^2 = (y-2)^2
16 - (-4)^2 = (y-2)^2
16 - 16 = (y-2)^2
0 = (y-2)^2
0 = y - 2
2 = y
⇒ C = (2,2)
Mel should use the least common multiple to solve the problem
<u>Solution:</u>
Given, Mel has to put the greatest number of bolts and nuts in each box so each box has the same number of bolts and the same number of nuts.
We have to find that should Mel use the greatest common factor or the least common multiple to solve the problem?
He should use least common multiple.
Let us see an example, suppose 12 bolts and nuts are to be fit in 6 boxes.
Then, if we took H.C.F of 12 and 6, it is 6, which means 6 bolts and nuts in each box, but, after filling 2 boxes with 6 bolts and nuts, there will be nothing left, which is wrong as remaining boxes are empty.
So the remaining method to choose is L.C.M.
Hence, he should use L.C.M method.
The histogram which shows the results of the time trial is option B
<h3>Histogram</h3>
Given data:
12:32, 12:46, 13:30, 15:03, 13:43, 13:46, 14:20, 14:29, 12:34, 14:43, 15:26, 16:02, 16:16, 14:38, 16:55, 17:10, 14:26, 17:21, 17:58, 13:30
Grouping of data,:
12:00 - 12:59 = 3
13:00 - 13:59 = 4
14:00 - 14:59 = 5
15:00 - 15:59 = 2
16:00 - 16:59 = 3
17:00 - 17:59 = 3
- The frequency is represented on the y-axis(number of cyclist)
- The time is represented on the x-axis
Learn more about histogram:
brainly.com/question/9388601
#SPJ1
