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Orlov [11]
3 years ago
11

triangle ABC has vertices at A (-12,-8) B (10,-6) and C (6,14. What is the midpoint of side AC of the triangle?

Mathematics
1 answer:
grigory [225]3 years ago
5 0

Step-by-step explanation:

midpoint of AC=(x¹+x²/2,y¹+y²/2)

=(-12+6/2,-8+14/2)

=(-6/2, 6/2)

=(-3,3)

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The following function represents the profit P(n), in dollars, that a concert promoter makes by selling tickets for n dollars ea
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A) zeroes

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P(n) = -250(n - 3)(n - 7)

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They rerpesent that if the promoter sells tickets at 3  or 7 dollars the profit is zero.

B) Maximum profit

Completion of squares

n^2 - 10n + 21 = n^2 - 10n + 25 - 4 = (n^2 - 10n+ 25) - 4 = (n - 5)^2 - 4

P(n) = - 250[(n-5)^2 -4] = -250(n-5)^2 + 1000

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C) Axis of symmetry

Vertex = (h,k) when the equation is in the form A(n-h)^2 + k

Comparing A(n-h)^2 + k with - 250(n - 5)^2 + 1000

Vertex = (5, 1000) and the symmetry axis is n = 5.



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3 years ago
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Answer:

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Divide 1342 by 100

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