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atroni [7]
1 year ago
9

The midpoint of AB is M (5, 6). If the coordinates of A are (3, 8), what are the

Mathematics
1 answer:
viktelen [127]1 year ago
3 0

Answer:

(7, 4)

Step-by-step explanation:

\text{Let the coordinates of B } = x_{B} , y_{B} \\\text{Let the coordinates of A} = x_{A} , y_{A} \\\text{Let the coordinates of midpoint M} = x_{M} , y_{M}

By the midpoint formula,
(x_{M},y_{M}) = (\frac{x_{A} + x_{B}}{2} , \frac{y_{A} + y_{B}}{2} )

x_{M} = \frac{x_{A} + x_{B}}{2} \\y_{M} = \frac{y_{A} + y_{B}}{2} \\

We have coordinates of midpoint as (5, 6) and coordinates of A as (3,8)
So
5 = \frac{3 + x_{B}}{2} \\\\\\textrm{Cross-multiplying } 10 = 3 + x_{B}} \textrm{ or }  x_{B} = 10-3 = 7\\

6 = \frac{8 + y_{B}}{2} \\\\\\\\textrm{Cross-multiplying } 12 = 8 + y_{B}} \textrm{ or }  y_{B} = 12-8 = 4\\

So coordinates of B are (7,4)

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Part 1) a=-\frac{1}{4c^6}

Part 2) a=-\frac{1}{4c^{-6}}

Step-by-step explanation:

Part 1) we have

a=2b^{3} ----> equation A

b=-\frac{1}{2}c^{-2} ----> equation B

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a=2(-\frac{1}{2}c^{-2})^{3}

Applying property of exponents

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therefore

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Part 2) we have

a=2b^{3} ----> equation A

b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2} ----> equation B

substitute equation B in equation A

a=2(-\frac{c^{2}}{2})^{3}

Applying property of exponents

(x^{m})^{n}=x^{m*n}

x^{-m} =\frac{1}{x^{m}}

a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})

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