The length of a rectangle is 2 cm longer than its width.

1 answer:

7 0

In a rectangle, there will be two sets of equal lengths:
56 = 2(x) + 2(x + 2)
56 = 2x + 2x + 4
56 = 4x + 4
52 = 4x
13 = x
Width = 13cm
Length = 15cm

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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:

<u>Given sequence</u>:

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

Calculate the <u>first differences</u> 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 <u>second differences:</u>

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

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

The <u>coefficient</u> of the n² term is <u>half of the second difference</u>.

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 <em>between</em> 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:

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