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navik [9.2K]
3 years ago
14

find the sum of a 9-term geometric sequence when the first term is 4 and the last term is 1,024 and select the correct answer be

low. 682 2,044 2,048 678
Mathematics
2 answers:
olasank [31]3 years ago
7 0
Hello,

u_{1} =4\\
 u_{2} =4*q\\
 u_{3} =4*q^2\\
...\\

 u_{9} =4*q^8\\\\
==\textgreater\ 4*q^8=1024\\
==\textgreater\ q^8=256\\
==\textgreater\ q=2\mbox{( and there is a problem in the question) } \ or \ q=-2\\
if\ q=2 \ then\  \sum_{i=0}^{8}\ 4*(2)^i= 4*\dfrac{1-2^9}{1-2} =2044\\

if \ q=-2 \ then\ \sum_{i=0}^{8}\ 4*(-2)^i= 4*\dfrac{1-(-2)^9}{1+2} =684\\





Answer B (but see the problem)
slavikrds [6]3 years ago
6 0

Answer:

I just did the test and the answer was 2,044.

Step-by-step explanation:

<h2>I Promise just trust me </h2>
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90,000 ..............
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3 years ago
The population of fish in a lake grew exponentially. In 2000, a researched estimated
cupoosta [38]

Answer:

a) P(t) = P₀ e⁰•⁰⁵⁴ᵗ

b) 2786 fishes

c) The estimated population will be 24000 in the year 2030.

Step-by-step explanation:

The function representing the population of fish in a lake as it grows exponentially

P(t) = P₀eᵏᵗ

If the base year is taken to be 2000

P₀ = 4780 fishes

k = constant

In 2010, P(t=10) = 8200 fishes, we can now solve for the constant

kt = k × 10 = 10k

8200 = 4780 e¹⁰ᵏ

e¹⁰ᵏ = (8200/4780) = 1.7155

In e¹⁰ᵏ = In 1.7155 = 0.5397

10k = 0.5397

k = 0.054 to 3 d.p

P(t) = P₀ e⁰•⁰⁵⁴ᵗ

b) What was the estimated population in 1990?

1990 is 10 years before 2000, hence, t = -10

P(t) = P₀ e⁰•⁰⁵⁴ᵗ

0.054t = 0.054 × -10 = -0.54

P₀ = 4780

P(t) = 4780 e⁻⁰•⁵⁴ = 4780 × 0.5829 = 2,786.4 = 2786 fishes to the nearest thousand

c) When will the population 24,000?

P(t) = 24000

P₀ = 4780

t = ?

24000 = 4780 e⁰•⁰⁵⁴ᵗ

e⁰•⁰⁵⁴ᵗ = (24000/4780) = 5.021

In e⁰•⁰⁵⁴ᵗ = In 5.021 = 1.6136

0.054t = 1.6136

t = (1.6136/0.054) = 29.88 years = 30 years to the nearest whole number.

Since the base year is 2000, 30 years after that is 2000 + 30 = 2030.

Hope this Helps!!!

8 0
3 years ago
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