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3 years ago

Find the midpoint of A and B where A has coordinates (2,9)

Mathematics

1 answer:

timofeeve [1]3 years ago

7 0

Answer:

$(\dfrac{9}{2}, \dfrac{13}{2})$

Step-by-step explanation:

The coordinates of A = (2,9)

The coordinates of B = (7,4)

Let M (x,y) be the mid point of A and B.

$M(x,y)=(\dfrac{x_1+x_2}{2}),(\dfrac{y_1+y_2}{2})\\\\=\dfrac{2+7}{2}, \dfrac{9+4}{2}\\\\=(\dfrac{9}{2}, \dfrac{13}{2})$

So, the coordinates of mid point of A and B is $(\dfrac{9}{2}, \dfrac{13}{2})$ .

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Alex

Answer:

Step-by-step explanation:

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3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$