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3 years ago

F(n)=64+6n what is the recursive formula of f(n)?

Mathematics

1 answer:

Ghella [55]3 years ago

5 0

Answer:

Step-by-step explanation:

f(n)=64+6n

f(n-1)=64+6(n-1)=64+6n-6=f(n)-6

f(n-1)+6=f(n)

or f(n)-f(n-1)=6

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Examine the following steps. Which do you think you might use to prove the identity Tangent (x) = StartFraction tangent (x) + ta

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Answer:

The correct options are;

1) Write tan(x + y) as sin(x + y) over cos(x + y)

2) Use the sum identity for sine to rewrite the numerator

3) Use the sum identity for cosine to rewrite the denominator

4) Divide both the numerator and denominator by cos(x)·cos(y)

5) Simplify fractions by dividing out common factors or using the tangent quotient identity

Step-by-step explanation:

Given that the required identity is Tangent (x + y) = (tangent (x) + tangent (y))/(1 - tangent(x) × tangent (y)), we have;

tan(x + y) = sin(x + y)/(cos(x + y))

sin(x + y)/(cos(x + y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y)) = (tan(x) + tan(y))(1 - tan(x)·tan(y)

∴ tan(x + y) = (tan(x) + tan(y))(1 - tan(x)·tan(y)

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Aliun [14]

The completed table is given below,

the percentage of girls in the class is 47.62%,

the percentage of boys in the class is (100-47.62)% = 52.38%,

the percentage of students that love Math is 36.5%,

the percentage of students that love Reading is 22%, and

the percentage of students that love Science is 41.27%

To make comparison and statistical analysis easier, numerical data is systematically and logically represented in tabulations as rows and columns. Grouping relevant data together makes comparison easier and aids in statistical analysis and interpretation.

To put it another way, tabulation is the process of organizing data and presenting it in a tabular format. Depending on the type of categorization, it might be double, complicated, or simple.

The below table yields that

The total number of students is 315

The fraction of girls in the class is 150/315 = 0.4762

So, the percentage of girls in the class is 47.62%

Hence the percentage of boys in the class is (100-47.62)% = 52.38%

The fraction of students that love Math is 115/315 = 0.365

So, the percentage of students that love Math is 36.5%

The fraction of students that love Reading is 70/315 = 0.22

So, the percentage of students that love Reading is 22%

The fraction of students that love Science is 130/315 = 0.4127

So, the percentage of students that love Science is 41.27%

Learn more about tabulation here-

brainly.com/question/10670417

#SPJ10

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