How do you evaluate $4 y^{2} + 5$ $4y ^ { 2} + 5$ when $y = 3$ $y=3$ ?

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How do you evaluate $4 y^{2} + 5$ $4y ^ { 2} + 5$ when $y = 3$ $y=3$ ?

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Answer 1 · 2018-03-04T13:37:15

$41$ $41$

Explanation:

$substitute y = 3 into the expression$ $"substitute y = 3 into the expression"$

$\Rightarrow (4 \times 3^{2}) + 5$ $rArr(4xxcolor(red)(3)^2)+5$

$= (4 \times 9) + 5 = 36 + 5 = 41$ $=(4xx9)+5=36+5=41$

Answer 2 · 2018-03-04T13:39:30

$= 41$ $= 41$

Explanation:

The first thing is to use the knowledge of squaring:

$y^{2} = y \times y$ $y^2 = y xx y$

Hence using "bidmas":

$4 y^{2} + 5 = (4 \times y \times y) + 5$ $4y^2 + 5 = (4xxyxxy) + 5$

We can just replace $y$ $y$ with $3$ $3$ :

$= (4 \times 3 \times 3) + 5$ $= (4xx3xx3) + 5$

$= (4 \times 9) + 5$ $= (4xx9) +5$

$= 36 + 5$ $= 36 + 5$

$= 41$ $= 41$

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