Let y be the unknown number
The other number is x - 12
Answer:
0.049168726 light-years
Step-by-step explanation:
The apparent brightness of a star is
where
<em>L = luminosity of the star (related to the Sun)
</em>
<em>d = distance in ly (light-years)
</em>
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
According to the inverse square law for light intensity
where
light intensity at distance
light intensity at distance
Let
be the distance we would have to place the 50-watt bulb, then replacing in the formula
Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.
Answer:
D
Step-by-step explanation:
I am sorry but I can not give you how I got thinms answer but i do know tgat the answer is correct
Answer:
Area of triangle = 144 sq.in
Area of rectangle = 312 sq.in
Area of whole figure = 456 sq.in
Step-by-step explanation:
area of truangle =
- 1/2 × 24 × 12
- 12 × 12
- 144
area of rectangle =
area of whole fig. = 312 + 144 = 456 sq.in
if you notice the triangle on the left-hand-side, it has a base of 8 ft and a height/altitude of 10.
now, the triangular pyramid has 3 of those triangles standing up, and one at the base lying down, so 4 triangles with that base and altitude. So we can simply find the area of all 4 of those triangles, sum them up and that's the area of the triangular pyramid.
![\bf \stackrel{\textit{area of the four triangles}}{4\left[ \cfrac{1}{2}(8)(10) \right]}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20four%20triangles%7D%7D%7B4%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%288%29%2810%29%20%5Cright%5D%7D%20)