What is the explicit rule for this geometric sequence? 4,45,425,4125,…

2 answers:

4 0

You did not write the question correctly but I will try to answer the question according to 2 possible interpretations of your question.

Case 1:

Given the geometric sequence:

$4,\ \frac{4}{5} ,\ \frac{4}{25} ,\ \frac{4}{125} ,\ .\ .\ .$

Common ratio is given by $r= \frac{2nd\ term}{1st\ term} = \frac{ \frac{4}{5} }{4} = \frac{1}{5}$

The explicit rule of a geometric sequence is given by $T_n=ar^{n-1}$

Therefore, the explicit rule of the given sequence is $T_n=4\left( \frac{1}{5} \right)^{n-1}\ or\ 20\left( \frac{1}{5} \right)^n$

Case 2:

Given the geometric sequence:

$4,\ 4\cdot5,\ 4\cdot25,\ 4\cdot125,\ .\ .\ .$

Common ratio is given by $r= \frac{2nd\ term}{1st\ term} = \frac{4\cdot5}{4} = 5$

The explicit rule of a geometric sequence is given by $T_n=ar^{n-1}$

Therefore, the explicit rule of the given sequence is $T_n=4(5)^{n-1}\ or\ \frac{4}{5}(5)^n$

zysi [14]3 years ago

4 0

Answer:

The Answer Is An=4⋅(1/5)n−1

Step-by-step explanation:

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Please help me I don’t understand this at all and I need to pass quits so please somebody out there help me

kvv77 [185]

Answer:

B) Undefined

Step-by-step explanation:

Vertical lines do not have a horizontal change; the line goes straight up and down. In other words, the change in the x values between any points on the graph is always 0, which would make the denominator of the slope 0. Since anything divided by 0 is undefined, the slope of a vertical line is <u>undefined</u>.

Source: https://www.brightstorm.com/math/algebra/linear-equations-and-their-graphs/finding-the-slope-of-a-line-from-a-graph-problem-3/

7 0

3 years ago

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0

4 years ago

WILL GIVE BRAINIEST what is the sum of the interior angles of the polygon shown below.

jolli1 [7]

Answer:

900

Step-by-step explanation:

(n-2)*180

5*180=900

7 0

3 years ago

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