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3 years ago

ASAP ANSWERRR If a sequence is defined recursively by f(0)=4 and f(n+1)= -3f(n)+1 for n≥0, the f(3) is equal to

Mathematics

1 answer:

LenaWriter [7]3 years ago

3 0

Idk maybe ask an adult

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Peter attends 6 dance lessons each week all year long a year has 52 weeks peter missed 5 dance lessons while sick how many dance

viktelen [127]

6 lessons per week multiplied by the number of weeks in a year
6 x 52 = 312
total - 5 missed
312 - 5 = 307 dance lessons

5 0

3 years ago

1 ln(1) + 2 ln(2) + 3 ln(3) + ⋯ + 10 ln(10)

Lapatulllka [165]

Answer:

$\ln(1\cdot2^{2} \cdot3^{3}\cdot4^{4} ...\cdot10^{10} )$

Step-by-step explanation:

By logarithm rules

$\ln1+2\ln2+3\ln3+....+10\ln10\\\ln1+\ln2^2+\ln3^3+....\ln10^{10}\\\ln(1\cdot2^{2} \cdot3^{3}\cdot4^{4} ...\cdot10^{10} )$

6 0

3 years ago

A factory made 900 jars of peanut butter. 35% of the jars contained creamy peanut butter. How many jars of creamy peanut butter

DIA [1.3K]

Answer:

315 jars

Step-by-step explanation:

This question is merely another way of writing: what is 35% of 900?

Now, when we look at it that way, it's easy to solve.

So, the way to solve is to either multiply 35% by 900, or just interpret that as 0.35 times 900. Anyway, the result is 315.

I hope this helped a bunch! Tell me if you need any further assistance...

( :

5 0

3 years ago

7(8+x+2) equals what exactly i dont know

Nataly [62]

Answer:

70 + 7x

Step-by-step explanation:

7(8+x+2)

Combine like terms in the parenthesis

7(8+2+x)

7(10+x)

Distribute

70 + 7x

4 0

3 years ago

Read 2 more answers

The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i

Anit [1.1K]

Answer:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The $n^{th}$ term of geometric sequence is given by:

$a_n = ar^{n-1}$

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}$

the first term can be calculated as:

$16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128$

Thus, the required geometric sequence is

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

4 0

3 years ago