The average distance between a star and Earth is typically measured in astronomical units (AU), which is the average distance between Earth and the Sun. To calculate the distance in AU, divide the star's distance in kilometers or miles by the average distance between Earth and the Sun.
The average distance from the Earth to a star is typically measured in astronomical units (AU). An astronomical unit is defined as the average distance between the Earth and the Sun, which is about 149.6 million kilometers or 93 million miles.
To calculate the distance to a star in AU, you need to determine the star's distance in kilometers or miles and then convert it to AU. For example, if a star is 900 million kilometers away, you divide that distance by the average distance between the Earth and the Sun to get the distance in AU. In this case, the star would be approximately 6 AU away from Earth.
It's important to note that the distances between stars and Earth are incredibly vast, so even a distance of a few AU is still very far.
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Answer:
11
Step-by-step explanation:
To find the 4th term, you simply put 4 in for n because n denotes which term:
The answer is thus 11.
Answer:
5/16
Step-by-step explanation:
x-3y=-11
substitution
Answer:
Amy is 1200 seconds fater than Bill.
Step-by-step explanation:
b. Use your variables to determine expressions for the volume, surface area, and cost of the can.
c. Determine the total cost function as a function of a single variable. What is the domain on which you should consider this function?
d. Find the absolute minimum cost and the dimensions that produce this value.
Answer:
a) file annex
b) V(c) = π*x²*y
A(x) = 2*π*x² + 32/x
C(x) = 0,1695*x² + 0,48 /x
Domain C(x) = {x/ x >0}
d) C(min) = 0,64 $
x = 1.123 in radius of base
y = 4,04 in height of the can
Step-by-step explanation:
See annex file
Lets:
call x = radius of the base of the cylinder and
y = the height of the cylinder
Then
Volume of the cylinder ⇒ V(c) = π*r²*h ⇒V(c) = π*x²*y
And y = V / ( π*x²) ⇒ V = 16 / ( π*x²)
Area of cylinder = lids area + lateral area
lids area = 2*π*x² ⇒ lateral area = 2*π*x*y
lateral area =2*π*x [16/(π*x²) ] ⇒ lateral area = 32/x
Then
A(x) = 2*π*x² + 32/x
Function cost C(x)
C(x) = 0.027 * 2*π*x² + 0.015 * (32/x)
C(x) = 0,1695*x² + 0,48 /x
Domain C(x) = {x/ x >0}
Now function cost:
C(x) = 0,1695*x² + 0,48 /x
Taking derivative:
C´(x) = 2*0,1695*x - 0.48/x² C´(x) = 0,339*x - 0.48/x²
C´(x) = 0 0.339*x³ - 0.48 = 0 x³ = 0.48/0.339 x³ = 1.42
x = 1.123 in
y = 16/πx² ⇒ y = 4,04 in
C(min) = 0,64 $