How do you evaluate $b = \sqrt{a^{2} - c^{2}}$ $b=sqrt(a^2-c^2)$ given a=29 and c=20?

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How do you evaluate $b = \sqrt{a^{2} - c^{2}}$ $b=sqrt(a^2-c^2)$ given a=29 and c=20?

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Answer 1 · 2016-11-24T11:56:10

$b = 35.23$ $b = 35.23$ rounded to the nearest 100th.

Explanation:

You substitute $29$ $29$ for $a$ $a$ and $20$ $20$ for $c$ $c$ in the equation and then do the math to solve for $b$ $b$ :

$b = \sqrt{29^{2} - 20^{2}}$ $b = sqrt(29^2 - 20^2)$

$b = \sqrt{29 \cdot 29 - 20 \cdot 20}$ $b = sqrt(29*29 - 20*20)$

$b = \sqrt{841 - 400}$ $b = sqrt(841 - 400)$

$b = \sqrt{1241}$ $b = sqrt(1241)$

$b = 35.23$ $b = 35.23$

Answer 2 · 2016-11-24T11:57:59

$b = 21$ $b=21$

Explanation:

To evaluate b, substitute the values given for a and c into the radical.

$\Rightarrow b = \sqrt{29^{2} - 20^{2}}$ $rArrb=sqrt(29^2-20^2)$

$= \sqrt{841 - 400} = \sqrt{441} = 21$ $=sqrt(841-400)=sqrt441=21$

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How do you evaluate $b = \sqrt{a^{2} - c^{2}}$ $b=sqrt(a^2-c^2)$ given a=29 and c=20?

2 Answers

Explanation:

Explanation:

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How do you evaluate $b = \sqrt{a^{2} - c^{2}}$ $b=sqrt(a^2-c^2)$ given a=29 and c=20?