Answer:
b(n) = -23 + 18n
Step-by-step explanation:
NOTE - This is an arithmetic sequence, so each number is increasing by addition. If you were in a geometric sequence, however, you'd be multiplying. This means that the explicit formula for the geometric sequence you'll be learning later on will be different.
The formula for explicit formula is b(n)=b(1)+d(n-1). The variable d represents the common difference and the a is whatever number of the term you are using.
Since you have b(1), you look at what your first term is. Plug -5 into a(1).
To find your common difference, find 13-(-5). This is 18, and adding it to the others you can see that it continues.
So, plugging in what you have so far it looks like b(n) = -5 + 18(n - 1)
All we need to do now is simplify.
1. Distribute: -5 + 18n - 18
2. Add like terms: -23 + 18n
Hope this helped and made sense lol.
Answer:
angels on a triangle = 180°
so you add 75 and 50 which gives you 125 then u take away that by 180 to find x (Just the process)
and that's your answer
It is asking you to find the sum of k^2 - 1 from k=1 to k=4. Since that is only 4 numbers, calculating the sum by hand wouldn’t be that bad.
(1^2 - 1) + (2^2 - 1) + (3^2 - 1) + (4^2 - 1) = 26
The easier way to find the sum is to use a few simple formulas.
When we have a term that is just a constant c, the formula is c*n.
When we have a variable k, the formula is k*n*(n+1)/2.
When we have a squared variable, the formula is k*n*(n+1)*(2n+1)/6.
In this case, we have a squared variable k^2 and a constant of -1.
So plug in n=4 to the formulas:
4*5*9/6 - 1*4 = 26
The answer is 26
Answer:
The polynomial -gh⁴i + 3g⁵ is a binomial, since it has two terms
Degree of polynomial: degree of a polynomial is the term with highest of exponent.
Degree of binomial -gh⁴i + 3g⁵ = 6
1st term(-gh⁴i ) = (power of g = 1, power of h = 4, power of i = 1)
2nd term(3g⁵) = (power of g = 5)
the polynomial -gh⁴i + 3g⁵ is a 6 degree binomial.
B
the other options all have a set price, the answers in b could all change price from month to month
