According to the given table- find the relation between x and y- x 1 2 3 4 y 2 6 12 20 1c A- y -2 x B- y -2 x -1C- y - x 2D- y-x2-x

The correct option is D y=x^2+xWhen x = 1, y = 1^2+1 = 1+1 = 2. When x = 2, y = 2^2 +2= 4+2=6. When x = 3, y = nbsp; 3^2 +3= 9+3=12. When x = 4, y = nbs ...

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Standard VII

Mathematics

Collection and Organisation of Data

According to ...

Question

According to the given table, find the relation between x and y.
$\begin{matrix} x & 1 & 2 & 3 & 4 y & 2 & 6 & 12 & 20 \end{matrix}$

y = 2 x

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y = 2 x - 1

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y = x^{2}

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y = x^{2} + x

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Solution

The correct option is D $y = x^{2} + x$
When x = 1, $y = 1^{2} + 1 = 1 + 1 = 2$ .

When x = 2, $y = 2^{2} + 2 = 4 + 2 = 6$ .

When x = 3, $y = 3^{2} + 3 = 9 + 3 = 12$ .

When x = 4, $y = 4^{2} + 4 = 16 + 4 = 20$ .

So, $y = x^{2} + x$

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According to the given table, find the relation between x and y. x1234y261220

The correct option is D y=x2+xWhen x = 1, y=12+1=1+1=2. When x = 2, y=22+2=4+2=6. When x = 3, y=32+3=9+3=12. When x = 4, y=42+4=16+4=20. So, y=x2+x

According to the given table, find the relation between x and y.
$\begin{matrix} x & 1 & 2 & 3 & 4 y & 2 & 6 & 12 & 20 \end{matrix}$

The correct option is D $y = x^{2} + x$
When x = 1, $y = 1^{2} + 1 = 1 + 1 = 2$ .

When x = 2, $y = 2^{2} + 2 = 4 + 2 = 6$ .

When x = 3, $y = 3^{2} + 3 = 9 + 3 = 12$ .

When x = 4, $y = 4^{2} + 4 = 16 + 4 = 20$ .

So, $y = x^{2} + x$