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2 years ago

What’s the midpoint of the line segment joining A and B A(2,-5);B(6,1)

1 answer:

7 0

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{6+2}{2}~~,~~\cfrac{1-5}{2} \right)\implies \left( \cfrac{8}{4}~,~\cfrac{-4}{2} \right)\implies (4,-2)$

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garik1379 [7]

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2 years ago

Read 2 more answers

How do you do this. Please give a full answer and explain why.

OverLord2011 [107]

Law of Exponent

$\displaystyle \large{ {a}^{ - n} = \frac{1}{ {a}^{n} } }$

Compare the terms.

$\displaystyle \large{ {a}^{ - n} = {( - 2)}^{ - 3} }$

Therefore, a = -2 and n = 3. From the law of exponent above, we receive:

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{ {( - 2)}^{ 3} } }$

Exponent Def. (For cubic)

$\displaystyle \large{ {a}^{3} = a \times a \times a }$

Factor (-2)^3 out.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{( - 2) \times ( - 2) \times ( - 2)}}$

(-2) • (-2) = 4 | Negative × Negative = Positive.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{4 \times ( - 2)}}$

4 • (-2) = -8 | Negative Multiply Positive = Negative.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{ - 8}}$

If either denominator or numerator is in negative, it is the best to write in the middle or between numerator and denominators.

Hence,

$\displaystyle \large \boxed{ {( - 2)}^{ - 3} = - \frac{1}{ 8}}$

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2 years ago

Given the equation Y -3 equals 1/2 (X +6 )in point slope form identify the equation of the same line in standard form

liberstina [14]

The standard form of a line:

$Ax+By=C$

We have the equation of a line in the slope-point form:

$y-3=\dfrac{1}{2}(x+6)$ use distributive property

$y-3=\dfrac{1}{2}x+3$ add 3 to both sides

$y=\dfrac{1}{2}x+6$ multiply both sides by 2

$2y=x+12$ subtract x from both sides

$-x+2y=12$ change the signs

$x-2y=-12$

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