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umka2103 [35]
2 years ago
14

PLEASE HELP ME! For a geometric sequence, if a1 = 2 and a5 = 1250, what is the value of r?​

Mathematics
1 answer:
Zina [86]2 years ago
5 0

Answer: r = 5

Steps:

a_n=ar^{n-1}

1. add Values to formula

1250=2r^{5-1}

2. get r alone

\frac{1250}{2}=\frac{2r^{4}}{2}  ➜ 625= r^4
3. get rid of the exponents

\sqrt[4]{625}= \sqrt[4]{r^4} ➜ \sqrt[4]{625}= r=5

r = 5

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You exercise for one hour each day and you burn 8.5 calories per minute. If you graphed the number of calories you burned as the
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Find the recursive rule, explicit rule, and f(20)
DiKsa [7]

Answer:

Recursive:

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Explicit:

f(n)=35+10(n-1)

And the 20th term is 225.

Step-by-step explanation:

We have the sequence:

35, 45, 55, 65.

Notice that each subsequent term is 10 more than the previous term.

Therefore, our common difference is (+)10.

Recursive Rule:

The standard format for the recursive rule is:

f(n)=a, f(n)=f(n-1)+d

Where a is the initial term and d is the common difference.

From our sequence, we know that a the initial term is 35.

And as determined, our common difference d is 10.

Substitute. Hence, our recursive rule is:

f(1)=35, f(n)=f(n-1)+10

Explicit Rule:

The standard format for the explicit rule is:

f(n)=a+d(n-1)

Where a is the initial term and d is the common difference. So, let’s substitute 35 for a and 10 for d. Hence, our explicit formula is:

f(n)=35+10(n-1)

Now, let’s find the 20th term. We will utilize the explicit rule since the recursive rule can get tedious. Substitute 20 for n because we would like to 20th term. Thus:

f(20)=35+10(20-1)

Evaluate:

\begin{aligned} f(20)&=35+10(19) \\ f(20)&=35+190 \\ f(20)&=225 \end{aligned}

Hence, the 20th term is 225.

5 0
2 years ago
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