M = 54 bro bro bro give me BRANLIEST
Step-by-step explanation:
LET THE Point 1 A and point 2 is B
(AB) ^2=(-9-(-3))^2+ (5-(-6))^2+(-4-7)^2=278
AB=16.67 length unit
Midpoint
X=-9-3/2 =-6
Y=5-6/2=-0.5
Z=-4+7/2=-1.5
Let:
Vj = Speed of jane
Vm = Speed of mike
d = distance
t = time
Jane travels 7 mph less than 2 times as fast as Mike, therefore:
Vj = 2Vm - 7
Remeber:
distance = speed*time
Distance traveled by mike:
d=Vm*t = Vm*6
Distance traveled by jane:
d + 198 = Vj*6
where:
Vj = 2Vm - 7
d + 198 = (2Vm-7)*6
Now, let:
d=Vm*6 (1)
d + 198 = (2Vm-7)*6 (2)
Replace (1) into (2)
6Vm + 198 = 12Vm - 42
Subtract 6Vm from both sides:
6Vm + 198 - 6Vm = 12Vm - 42 - 6Vm
198 = 6Vm - 42
Add 42 to both sides:
198 + 42 = 6Vm - 42 + 42
240 = 6Vm
Divide both sides by 6:
240/6 = 6Vm/6
40 = Vm
Vm = 40mph
Replace Vm into this equation Vj = 2Vm - 7 :
Vj = 2(40) - 7 = 80 - 7 = 73mph
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Explanation:
Here's one way to go at it.
Draw segments AB and CO. Define angles as follows. (The triangles with sides that are radii are all isosceles, so their base angles are congruent.)
x = angle OAB = angle OBA
y = angle OAC = angle OCA
z = angle OBC = angle OCB
Consider the angles at each of the points A, B, C.
At A, we have ...
angle CAB = x + y
At B, we have ...
angle CBA = x + z
At C, we have ...
angle ACB = y + z
The sum of the angles of triangle ABC is 180°, as is the sum of angles in triangle ABO. This gives ...
x + x + ∠AOB = (x+y) +(x+z) +(y+z)
∠AOB = 2(y+z) = 2∠ACB
This shows ∠AOB = 2×∠C, as required.
The answer to the question is B (the second option 4:2)