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mariarad [96]
3 years ago
6

X+2y=4 -x-2y=-4the solutions ​

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
6 0

subtract equations to form the equation of:

2x + 2y = 8

rearrange first equation to make x the subject:

x = 4 - 2y

subsitute into 2x + 2y = 8

2 (4 - 2y) + 2y = 8

8 - 4y + 2y = 8

collect like terns

-2y + 8 = 8

8s cancel out

-2y = 0

divide by -2

<u>y</u><u> </u><u>=</u><u> </u><u>0</u>

substitute y value into x + 2y = 4

x + 2 (0) = 4

solve for x

x + 0 = 4

<u>x</u><u> </u><u>=</u><u> </u><u>4</u>

therefore, the solutions are:

<h3><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>4</u></h3><h3><u>y</u><u> </u><u>=</u><u> </u><u>0</u></h3>

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Answer:

y = −3x+7

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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}

Finally, we can clearly see that the operation to get from 3n² + 5n to the given sequence is to subtract 2.

Therefore, the nth term of the quadratic sequence is:

3n² + 5n - 2

6 0
1 year ago
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