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3 years ago

The midpoint of a segment is (6,−4) and one endpoint is (13,−2). Find the coordinates of the other endpoint.

Mathematics

1 answer:

KengaRu [80]3 years ago

8 0

Answer: A. (-1, -6)

Step-by-step explanation:

Use the midpoint formula:

Endpoint #1 = (x₁, y₁) = (13, -2)
Endpoint #2 = (x₂, y₂)

$midpoint = (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) \\\\(6, -4) = (\frac{13+x_{2}}{2}, \frac{-2+y_{2}}{2})\\\\\frac{13+x_{2}}{2} =6\\\\13+x_{2}=6*2\\\\x_{2}=12-13=-1\\\\ \\ \frac{-2+y_{2}}{2}=-4\\\\-2+y_{2}=(-4)*2\\\\y_{2}=-8+2=-6\\\\\\\left \{ {{x_{2}=-1} \atop {y_{2}=-6}} \right.$

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Elodia [21]

Answer:

Monthly Payment Amount for Millie = $223.73

Step-by-step explanation:

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Cashback offered by the dealer = $2700

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4 years ago

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Oduvanchick [21]

Answer:

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Step-by-step explanation:

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4 years ago

(7, 5) and (-3, 4) written in slope intercept form

Rom4ik [11]

Answer:

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Step-by-step explanation:

y - y1/ x - x1 = y2 - y1/ x2 - x1

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4 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

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