C(-8,2) and M(0,0) , since M is at the origin. Let x₁ and y₁ be the
coordinates of S →s(x₁ , y₁)
C(-8,2) and S(x₁ , y₁)
The coordinates of M, the midpoint of CS are M(x₂ , y₂)
a) x₂ = (-8 + x₁)/2 , but x₂ = 0, then :
0 = -4+x₁/2 and x₁ = 8
b) y₂ = (2+y₁)/2 , but y₂ = o, then:
0 = 2+ y₁/2 and y₂ = -2
Then the coordinates of S are S(8 , -2)
Answer:
no
Step-by-step explanation:
The given lengths do not satisfy the Pythagorean theorem:
34² ≠ 33² +16²
1156 ≠ 1089 +256
The given sides do not form a right triangle.
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<em>Alternate solution</em>
You can also compute 34² -33² = (34+33)(34-33) = 67·1 = 67 ≠ 16².
Even simpler, a Pythagorean triple will never have zero or two even numbers. The numbers (16, 33, 34) cannot be a Pythagorean triple.
Number of tickets: T.
Number of customers: c
Initially the number of tickets is T0=150, when the group hasn't sold any tickets (c=0). Then the graph must begin with c=0 and T=150. Point=(0,150). Possible options: Graph above to the right and graph below to the left.
They sell the tickets in pack of three tickets per customer c, then each time they sell a pack of three tickets to a customer, the number of tickets is reduced by 3 (-3c). Then the number of tickets, T, the group has left after selling tickets to c customers is:
T=150-3c→T=-3c+150
For T=0→-3c+150=0→150=3c→150/3=c→c=50. The graph must finish with c=50, T=0. Final point=(c,T)=(50,0)
Answer:
The correct graph is above to the right, beginning on vertical axis with T=150 and finishing on horizontal axis with c=50.
The correct equation is T=-3c+150
If P=360, then 360yd / 4 sides = 90yds each side. Now that it is a rectangle, the Length should be more than the width, so if the length is 20yds more than the width, do 90 - 20 two times, Width= [ (90yds - 20yds)*2 ]
To sum things up, Length = 110; Width = 70. | 110*2 = 220, 70 * 2 = 140, 220+140 = 360.
There is nothing there sorry