5p2+q2+r2−−−−−−−−−−−√5p2+q2+r2\sqrt {5{p^2} + {q^2} + {r^2}} is equal to -" /> 5p2+q2+r2−−−−−−−−−−−√5p2+q2+r2\sqrt {5{p^2} + {q^2} + {r^2}} is equal to -" /> 5p2+q2+r2−−−−−−−−−−−√5p2+q2+r2\sqrt {5{p^2} + {q^2} + {r^2}} is equal to -" />

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If p : q : r = 1 : 2 : 4, then $\sqrt{5 p^{2} + q^{2} + r^{2}}$ $\sqrt {5{p^2} + {q^2} + {r^2}}$ is equal to -

Answer:- 1

Explanation:-

Solution:

\begin{aligned} p = k, q = 2 k, r = 4 k . \\ Then, \\ \sqrt{5 p^{2} + q^{2} + r^{2}} \\ = \sqrt{5 k^{2} + {(2 k)}^{2} + {(4 k)}^{2}} \\ = \sqrt{5 k^{2} + 4 k^{2} + 16 k^{2}} \\ = \sqrt{25 k^{2}} \\ = 5 k \\ = 5 p . \end{aligned}

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is equal to -", "text": "If p : q : r = 1 : 2 : 4, then

\sqrt{5 p^{2} + q^{2} + r^{2}}

$\sqrt {5{p^2} + {q^2} + {r^2}}$ is equal to -", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} p = k, q = 2 k, r = 4 k . \\ Then, \\ \sqrt{5 p^{2} + q^{2} + r^{2}} \\ = \sqrt{5 k^{2} + {(2 k)}^{2} + {(4 k)}^{2}} \\ = \sqrt{5 k^{2} + 4 k^{2} + 16 k^{2}} \\ = \sqrt{25 k^{2}} \\ = 5 k \\ = 5 p . \end{aligned}

$\eqalign{ & p = k,\,q = 2k,\,r = 4k. \cr & {\text{Then,}} \cr & \sqrt {5{p^2} + {q^2} + {r^2}} \cr & = \sqrt {5{k^2} + {{\left( {2k} \right)}^2} + {{\left( {4k} \right)}^2}} \cr & = \sqrt {5{k^2} + 4{k^2} + 16{k^2}} \cr & = \sqrt {25{k^2}} \cr & = 5k \cr & = 5p. \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: