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Lady_Fox [76]
3 years ago
9

Find a polynomial function f(n) such that f(1), f(2), ... , f(8) is the following sequence.

Mathematics
1 answer:
Irina18 [472]3 years ago
5 0
Observing there is a difference of +8 between each f(x).
The slope of this sequence is 8.(y2-y1)/(x2-x1)
Using the point slope equation,
y-y1 = m(x-x1) yields

y = 8x -4 as the answer.
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Given the points A(3, 4) and B (6, 10), find the
Dvinal [7]

Answer:

P (5 , 8)

Step-by-step explanation:

P (x,y) partition A (x₁ , y₁) B (x₂ , y₂) into ratio AM:MB = a:b = 2:1  ... a=2 , b=1

x = (bx₁ + ax₂) / (a+b)

  = (1 * 3 + 2 * 6) / (2 + 1)

  = 15/3

  = 5

y = (by₁ + ay₂) / (a+b)

  = (1 * 4 + 2 * 10) / (2 + 1)

  = 24/3

  = 8

P (5 , 8)

8 0
3 years ago
A geometric sequence is shown below.
mestny [16]

Answer:

An explicit representation for the nth term of the sequence:

f_n=-\frac{1}{2}\cdot \:4^{n-1}

It means, option (B) should be true.

Step-by-step explanation:

Given the geometric sequence

-\frac{1}{2},\:-2,\:-8,\:-32,...

A geometric sequence has a constant ratio, denoted by 'r', and is defined by

f_n=f_1\cdot r^{n-1}

Determining the common ratios of all the adjacent terms

\frac{-2}{-\frac{1}{2}}=4,\:\quad \frac{-8}{-2}=4,\:\quad \frac{-32}{-8}=4

As the ratio is the same, so

r = 4

Given that f₁ = -1/2

substituting r = 4, and f₁ = -1/2 in the nth term

f_n=f_1\cdot r^{n-1}

f_n=-\frac{1}{2}\cdot \:4^{n-1}

Thus, an explicit representation for the nth term of the sequence:

f_n=-\frac{1}{2}\cdot \:4^{n-1}

It means, option (B) should be true.

8 0
2 years ago
Explain the steps involved in adding two rational<br> expressions.<br> DONE
Helen [10]

Answer

Step-by-step explanation: example

5/(X+2). + 2/(X+1)

Common denominator. Is both (X+2)(X+1)

So the first fraction needs (X+1) since it already has (X+2) the second fraction needs ( X+2)

5(X+1). / (X+2)((X+1) +. 2/(X+2)/(X+1)

So multiply 5 with X then 5 with 1 and 2 times X and 2 times 2

5x+5 + 2x+4 / (X+2)((X+1)

Answer 7x+9/ (X+2)(X+1)

8 0
3 years ago
Tan x-1/tan x+1=1-cot x/1+cotx
ikadub [295]
\dfrac{\tan x-1}{\tan x+1}=\dfrac{\frac{\sin x}{\cos x}-1}{\frac{\sin x}{\cos x}+1}=\dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{1-\frac{\cos x}{\sin x}}{1+\frac{\cos x}{\sin x}}=\dfrac{1-\cot x}{1+\cot x}
6 0
3 years ago
What is the value of the expression? 10(5 + 16) = (3+4)
Kamila [148]
Well it’s 210 = 7 and it’s false
7 0
2 years ago
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