Question

10.
The relation 7piR3t = rt3+6 describes the position of a particle in one direction where x is in metres
and in sec. The displacement, when velocity is zero, is
(A) 24 metres (B) 12 metres
(C) 5 metres
(D) Zero​

Answer 1

Answer:

4(a)

\begin{gathered}{ \huge{ \blue{ \underline{ \underline{ \bf{ \red{ \mathtt{question}}}}}}}} \\ \\ 125 - [ -20 \: of \: (-16) + {15 ÷ (-5)}] \\ \\ { \huge{ \blue{ \underline{ \underline{ \bf{ \red{ \mathtt{solution}}}}}}}}\end{gathered}

question

125−[−20of(−16)+15÷(−5)]

solution

125 - [ -20 × -16 + {-3}]

125 - [320 + (-3)]

125 - [ 320 - 3]

125 - 317

- 192.

4(b)

\begin{gathered}{ \huge{ \blue{ \underline{ \underline{ \bf{ \red{ \mathtt{question}}}}}}}} \\ \\ [180 ÷ {12×10 - 15 ×9}] + 2 \\ \\ { \huge{ \blue{ \underline{ \underline{ \bf{ \red{ \mathtt{solution}}}}}}}}\end{gathered}

question

[180÷12×10−15×9]+2

solution

[180 ÷ {12×10 - 15 ×9}] + 2

[180 ÷ { 120 - 135}] + 2

[180 ÷ ( - 15)] + 12

-12 + 12

0

Hope it helps......

\huge \boxed{ \mathbb{ \fcolorbox{orange}{red}{thanks}}}

thanks