Part A
After dividing the first two terms by the coefficient of n², the coefficient of the linear term is -6, so we can complete the square by adding (and subtracting) the square of half that: (-6/2)² = 9.
... g(n) = n² -6n + 9 + 16 - 9
... g(n) = (n -3)² +7 . . . . . . . rewrite to vertex form
Part B
The generic vertex form is
... y = a(x -h)² +k . . . . . . for vertex (h, k) and vertical expansion factor "a"
Comparing this to g(n), we see a=1, h=3, k=7. When a > 0, the parabola opens upward, and the vertex is a minimum. Here, we have a > 0, so we can conclude ...
... the vertex (3, 7) is a minimum
Part C
The axis of symmetry is the vertical line through the vertex.
... x = 3
Answer:
So first you make a expression for the length and width.
(a^2+3)x(a^2+2a+5)
(2a+3)x(4a+5)
Combine like terms
(2a+4a)x(3+5)
6ax8
This is the area
Step-by-step explanation:
Answer:
1) yes--common difference is 3.
The line and parabola intersect at about x=-2 and x=1.5 so the solution is A. x = -2
Answer is 8671/6 which is the third choice
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Work Shown:
Find the first term of the sequence by plugging in n = 1
a_n = (5/6)*n + 1/3
a_1 = (5/6)*1 + 1/3 replace n with 1
a_1 = 5/6 + 1/3
a_1 = 5/6 + 2/6
a_1 = 7/6
Repeat for n = 58 to get the 58th term
a_n = (5/6)*n + 1/3
a_58 = (5/6)*58 + 1/3 replace n with 58
a_58 = (5/6)*(58/1) + 1/3
a_58 = (5*58)/(6*1) + 1/3
a_58 = 290/6 + 1/3
a_58 = 145/3 + 1/3
a_58 = 146/3
Now we can use the s_n formula below with n = 58
s_n = (n/2)*(a_1 + a_n)
s_58 = (58/2)*(a_1 + a_58) replace n with 58
s_58 = (58/2)*(7/6 + a_58) replace a_1 with 7/6
s_58 = (58/2)*(7/6 + 146/3) replace a_58 with 146/3
s_58 = (58/2)*(7/6 + 292/6)
s_58 = (58/2)*(299/6)
s_58 = (58*299)/(2*6)
s_58 = 17342/12
s_58 = 8671/6