Choose the correct transformation.

Use the function notation to write a recursive formula to represent the sequence: 4, 8, 12, ...

melamori03 [73]

Answer:

The correct answer is f(n) = f(n − 1) + 4

Step-by-step explanation:

Let us consider the sequence: 4, 8, 12, …

Using f(n) = f(n − 1) + 4, we can easily get the sequence.

As the first term is 4 and next term is obtained by adding 4 to the first term.

i.e. 4+4 = 8

8+4 = 12

12+4 = 16 and so on.

Since the next term is obtained by adding 4 to the previous term.

So, f(n) = f(n-1) + 4 would be the correct recursive formula for the function of the sequence 4, 8, 12, ....

Verification:

f(n) = f(n-1) + 4

Putting n=2 in f(n) = f(n-1) + 4 to get the second term of the sequence.

f(2) = f(2-1) + 4

f(2) = f(1) + 4

f(2) = 4 + 4

f(2) = 8

Putting n=3 in f(n) = f(n-1) + 4 to get the third term of the sequence.

f(3) = f(3-1) + 4

f(3) = f(2) + 4

So, adding 4 in f(2)=8 would give us the next term i.e. 12

f(3) = 8 + 4

f(3) = 12

Keywords: recursive formula, sequence

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Answer:

168cm is the right answer

Step-by-step explanation:

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Mrs. Gomes found that 40% of students at her high school take chemistry. She randomly surveys 12 students. What is the probabili

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Binomial formula is :
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Answer: C ) 0.213

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Round 12.5478 to the nearest hundredth's

sp2606 [1]

Answer:

I'm pretty sure it's 12.55

Step-by-step explanation:

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1 point) Are the functions f,g, and h given below linearly independent? f(x)=e3x+cos(5x), g(x)=e3x−cos(5x), h(x)=cos(5x). If the

Fynjy0 [20]

Answer:

Functions are linearly dependent (are not linearly independent.)

Step-by-step explanation:

Remember that two functions f(x), g(x) and h(x) are said linearly independent on an interval I if the only solution to the equation

$\alpha f(x)+\beta g(x)+\omega h(x)=0, \ \text{for all } x\in I$

is the trivial one: α = 0, β = 0, ω = 0. If they are not linearly independent, they are called linearly dependent.

Now, let f(x), g(x) and h(x) be the functions:

$f(x)=e^{3x}+\cos(5x),$

$g(x)=e^{3x}-\cos(5x),$

$h(x)=\cos(5x).$

Then, letting α = 1, β= -1 and ω = -2, we see that:

$\alpha f(x)+\beta g(x)+ \omega h(x)=e^{3x}+\cos(5x)-e^{3x}+\cos(5x)+2\cos(5x)=0.$

Hence, the functions f(x), g(x) and h(x) are not linearly independent, or equivalently, are linearly dependent.

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