Question

send me
$the \: answer \: when \: we \: send$
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Answer 1

Answer:

hi ok i will do

Explanation:

Answer 2

A N S W E R :

Volume of the milk container is 3080 cm³.

$\underline{\underline{\sf \bigstar\qquad Given :\qquad \bigstar}}$

$\frak{\pink{We \: have}}\begin{cases} \textsf{\red{Base diameter of cylindrical container = \textbf{14 cm}}}\\ \textsf{\purple{Height of the cylindrical container = \textbf{ 20 cm}}}\end{cases}$

$\underline{\underline{\sf \bigstar\qquad To \:Find:\qquad \bigstar}}$

We have to find the volume of the cylindrical container.

$\underline{\underline{\sf \bigstar\qquad Solution : \qquad \bigstar}}$

$\ddag \: \underline{\frak{To \: find \: the \: volume \: of \: cylindrical \: tank \: first \: we \: have \: to \: find \: the \: base \: radius \: of \: cylindrical \: \: container}} \: \ddag$

$:\implies \sf Base \: radius \: of \: cylindrical \: container = \dfrac{Base \: diameter \: of \: cylindrical \: container}{2}$

$:\implies \sf Base \: radius \: of \: cylindrical \: container = \dfrac{14}{2}$

$:\implies \underline{ \boxed{\sf Base \: radius \: of \: cylindrical \: container =7 \: cm}}$

$\therefore\:\underline{\textsf{The base radius of cylindrical container is \textbf{7 cm}}}.$

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$\ddag \: \underline{\frak{Now, let's \: find \: the \: volume \: of \: cylindrical \: container :}} \: \ddag$

$\dashrightarrow\:\:\sf Volume \: of \: cylindrical \: container = \pi r^2 h$

$\dashrightarrow\:\:\sf Volume \: of \: cylindrical \: container = \dfrac{22}{7} \times 7 \times 7 \times 20$

$\dashrightarrow\:\:\sf Volume \: of \: cylindrical \: container = 22\times 7 \times 20$

$\dashrightarrow\:\: \underline{ \boxed{\sf Volume \: of \: cylindrical \: container = 3080 \: {cm}^{3}}}$

$\therefore\:\underline{\textsf{The volume of cylindrical container is \textbf{3080 cm ^{\text3} }}}.$

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$\underline{\underline{\sf \bigstar\qquad Some \: important \: formulas \: related \: to \: it :\qquad \bigstar}}$

$\boxed{\bigstar{\sf \ Cylinder :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cylinder= \pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ cylinder= 2\pi r h\\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ cylinder= 2\pi r (h+r)$

$\boxed{\bigstar{\sf \ Cone :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cone= \dfrac{1}{3}\pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Cone = \pi r l \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Cone = \pi r (l+r) \\ \\ \\ \sf {\textcircled{\footnotesize4}} Slant \ Height \ of \ cone (l)= \sqrt{r^2+h^2}$

$\boxed{\bigstar{\sf \ Hemisphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Hemisphere= \dfrac{2}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Hemisphere = 2 \pi r^2 \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Hemisphere = 3 \pi r^2$

$\boxed{\bigstar{\sf \ Sphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Sphere= \dfrac{4}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Surface\ Area \ of \ Sphere = 4 \pi r^2$