The recursive formula of the geometric sequence is given by option D; an = (1) × (5)^(n - 1) for n ≥ 1
<h3>How to determine recursive formula of a geometric sequence?</h3>
Given: 1, 5, 25, 125, 625, ...
= 5
an = a × r^(n - 1)
= 1 × 5^(n - 1)
an = (1) × (5)^(n - 1) for n ≥ 1
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Answer:
z=-3 so 2y+3(-3)=-8, 2y-9=-8, add 9 to both sides 2y=1 divide both sides by 2, y=1/2, so 3x+4(1/2)+-3=-2, 3x+2-3=-2, 3x-1=-2, add 1 to both sides, 3x=-1, divide both sides by 3, x=-1/3
4^(12-4x)=256 realize that 256 is 4^4 so we really have:
4^(12-4x)=4^4 taking the natural log of both sides
(12-4x)ln4=4ln4 dividing both sides by ln4
12-4x=4 subtract 12 from both sides
-4x=-8 divide both sides by -4
x=2
Answer:
Answers are below!
Step-by-step explanation:
(2 + g) (8)
= (2 + g) (8)
Add a 8 after the 2, and flip.
= (2)(8) + (g)(8)
= 16 + 8g
= 8g + 16
= (4) (8 + -5g)
Add another 4, then flip.
= (4) (8) + (4) (-5g)
= 32 − 20g
= - 20g + 32
−7 (5-n)
= (−7) (5 + -n)
Add another 7, then flip.
= (−7) (5) + (-7) (-n)
= −35 + 7n
= 7n - 35
Use the distributive property.
a (b + c) = ab + ac
a = 8
b = 2m
c = 1
= 8 × 2m + 8 × 1
Simplify, you get 16m + 8.
Use the distributive property.
a (b + c) = ab + ac
a = 6x
b = y
c = z
= 6xy - 6xz is the answer.
Apply minus plus rules.
Multiply the numbers.
3 x 2 = 6