Answer:
n squared + 3n + 1
Step-by-step explanation:
5,11,19,29
Firstly look at the difference between each number. The first difference is 6 then 8 then 10 etc. After that you look at your created sequence - 6,8,10 etc. The difference is 2 each time. Then applying rules, you have to do the constant difference divided by 2 to get a coefficient of n squared. So in this case it's n squared because 2/2 = 1 so you don't have to place a 1 in front of the n squared. After you create a sequence from the n squared. That would be 1,4,9 etc. Then you need to see how to get from the sequence: 1,4,9 etc to your original sequence: 5,11,19 etc. So if you calculate it you will get 4,7,10 because firstly 5-1 = 4 then 11-4 = 7 etc. The sequence 4,7,10 is a linear sequence so the constant difference is 3 each time. So to get a nth term of a linear sequence you will start off as 3n then you will substitute 1 then 2 then 3 into the 3n. Therefore that would be 3,6 etc. So if you take the first substituted term, that would be 3 as said before then you will have to see how to get from the 3 to 4 so that is just adding 1. So the nth term of this linear sequence is 3n + 1. Check if it works at the end. So the overall nth term of the quadratic sequence is n squared as said before + 3n + 1.
Answer:
0
Step-by-step explanation:
Put -6 where x is and do the arithmetic.
f(-6) = 1/3 (-6 +6) = 1/3 (0)
f(-6) = 0
Answer:
A = 189 sq units
Step-by-step explanation:
You can cut the shape up into pieces. See image. There are two rectangles and a triangle. Some sides are given:
rectangle: 16 × 8
sm rectangle: 3 × 4
Some sides you have to add to find (the height of the whole entire thing: 16+4+4)
Some lengths you have to subtract to find (height of the triangle: 24 - 10).
Area of a rectangle:
length × width OR
base × height
Area of a triangle:
1/2 • b • h
Add up all three areas. See image.
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.
