. Which Statement Is True??

Mathematics

1 answer:

Zepler [3.9K]2 years ago

6 0

C bc it’s greater that the original

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(6t^4-t+1) - (6t^3-4t)

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Answer:

$= 6 {t}^{4} - 6 {t}^{3} + 3t + 1$

Step-by-step explanation:

$6 {t}^{4} - t + 1 - 6 {t}^{3} + 4t \\ = 6 {t}^{4} - 6 {t}^{3} + 3t + 1$

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What is the nth term rule of the quadratic sequence below?

Vladimir [108]

Answer:

3n² + 5n - 2

Step-by-step explanation:

Given sequence:

6, 20, 40, 66, 98, 136, ...

Calculate the first differences between the terms:

$6 \underset{+14}{\longrightarrow} 20 \underset{+20}{\longrightarrow} 40 \underset{+26}{\longrightarrow} 66 \underset{+32}{\longrightarrow} 98 \underset{+38}{\longrightarrow} 136$

As the first differences are not the same, calculate the second differences:

$14 \underset{+6}{\longrightarrow} 20 \underset{+6}{\longrightarrow} 26 \underset{+6}{\longrightarrow} 32 \underset{+6}{\longrightarrow} 38$

As the second differences are the same, the sequence is quadratic and will contain an n² term.

The coefficient of the n² term is half of the second difference.

Therefore, the n² term is: 3n²

Compare 3n² with the given sequence:

$\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 & 3 & 12 & 27 & 48 \\\cline{1-5} \sf operation & +3&+8 & +13 & +18 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}$

The second operations are different, therefore calculate the differences between the second operations:

$3 \underset{+5}{\longrightarrow} 8 \underset{+5}{\longrightarrow} 13\underset{+5}{\longrightarrow} 18$

As the differences are the same, we need to add 5n as the second operation:

$\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 +5n & 8&22 & 42 & 68\\\cline{1-5}\sf operation & -2 &-2 &-2 & -2 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}$