Question

d. Because it involves a
3. Kartik is packing juice packs in cartons such that each carton has 40 juice packs. After
packing j cartons, he is left with 25 juice packs. Which of these expressions
represents the total number of juice packs?
a. 40j + 25
b. 40j - 25
c. 25j + 40
d. 25j - 40​

Answer 1

Answer:

xhdjsjs

Step-by-step explanation:

djsjsisdssjsjzjuxdvsuxff

Answer 2

Answer:

$\large{\mathbb{\colorbox{purple} {\boxed{\boxed{\colorbox{white} {-:Answer:-}}}}}} </p><p>$

$</p><p></p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>\large{ \pmb{ \underline{ \underline{\frak{ \color{peru}{Given::}}}}}}</p><p>$

$\pink{➠}{ \sf{ \bf{ \frac {dv}{dt}} = \sf1 {cm}^{3}/s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\: \: \: \: \: \: \: ...(i) }}➠ </p><p>dt</p><p>dv</p><p> </p><p>$

$</p><p>\large{ \pmb{ \underline{ \underline{\frak{ \color{pink}{To \: find::}}}}}}</p><p>$

$.\pink{➠}{ \sf{ Rate \: of \: decrease \: of \: slant \: height.}}.$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{pin}{Formula \: used::}}}}}}$

$</p><p>\pink{➠}{ \sf{ Volume_{(cone)}= \frac{1}{3}\pi {r}^{2}h }}$

\large{ \pmb{ \underline{ \underline{\frak{ \color{plum}{Construction::}}}}}}

Construction::

Construction::

\pink{➠}{ \sf{ Kindly \: refer \: the \: attachment\:also!!}}➠Kindlyrefertheattachmentalso!!

\large{ \pmb{ \underline{ \underline{\frak{ \color{blue}{Concept \: required::}}}}}}

Conceptrequired::

Conceptrequired::

\pmb{ \bf{From \: the \: above \: attachment..}}

Fromtheaboveattachment..

Fromtheaboveattachment..

\pink{➠}{ \sf{cos \: {\bf x} =\frac{h}{l} = \frac{ \sqrt{3} }{2} }}➠cosx=

l

h

=

2

3

\pink{➠}{ \sf{Radius_{(cone)},(r)= \frac{1}{2} }}➠Radius

(cone)

,(r)=

2

1

\pink{➠}{ \sf{Height_{(cone)},(h)= \frac{ \sqrt{3} }{2} l}}➠Height

(cone)

,(h)=

2

3

l

\large{ \pmb{ \underline{ \underline{\frak{ \color{cyan}{According \: to \: Question::}}}}}}

AccordingtoQuestion::

AccordingtoQuestion::

\pmb{ \bf{Let's \: start \: with \: help \: of \: formulas!!}}

Let

′

sstartwithhelpofformulas!!

Let

′

sstartwithhelpofformulas!!

\pink{➠}{ \sf{ Volume_{(cone)}= \frac{1}{3}\pi {r}^{2}h }}➠Volume

(cone)

=

3

1

πr

2

h

{: : \implies{ \sf{ Volume_{(cone)}= \frac{1}{3}\pi { \big( \frac{1}{2} \big)}^{2} \Big( \frac{ \sqrt{3} }{2}l \Big) }}}::⟹Volume

(cone)

=

3

1

π(

2

1

)

2

(

2

3

l)

{: : \implies{ \sf{ Volume_{(cone)}= \frac{\pi}{8 \sqrt{3} } }}}::⟹Volume

(cone)

=

8

3

π

\pink{➠}{ \sf{ \bf{ \frac {dv}{dt}} = \sf1 {cm}^{3}/s \: \: \: \: \: \: \:\{ from \: eq. \: ..(i) \}}}➠

dt

dv

=1cm

3

/s{fromeq...(i)}

: : \implies{ \sf{ \bf{ \frac {d}{dt}}} \Big[\frac{\pi}{8 \sqrt{3}} {l}^{3} \Big ] = \sf1 {cm}^{3}/s }::⟹

dt

d

[

8

3

π

l

3

]=1cm

3

/s

: : \implies{ \sf \frac{\pi}{8 \sqrt{3}} { \bf{ \frac {d}{dt}}}{(l)}^{3} = \sf1 {cm}^{3}/s }::⟹

8

3

π

dt

d

(l)

3

=1cm

3

/s

: : \implies{ \sf \frac{3 {l}^{2}\pi }{8 \sqrt{3}} { \bf{ \frac {d}{dt}}} = \sf1 {cm}^{3}/s }::⟹

8

3

3l

2

π

dt

d

=1cm

3

/s

{: : \implies{ \sf \frac{ \sqrt{3} (16\pi )}{8 } { \bf{ \frac {dl}{dt}}} = \sf1 {cm}^{3}/s \: \: \: \: \: \: \: \: \{∵At \: l=4cm \}}}::⟹

8

3

(16π)

dt

dl

=1cm

3

/s{∵Atl=4cm}

{: : \implies{ \sf2 \sqrt{3} \pi { \bf{ \frac {dl}{dt}}} = \sf1 {cm}^{3}/s}}::⟹2

3

π

dt

dl

=1cm

3

/s

{: : \implies{ \sf{ \bf{ \frac {dl}{dt}}} = \sf \: \frac{1}{2 \sqrt{3}\pi } {cm}^{3}/s}}::⟹

dt

dl

=

2

3

π

1

cm

3

/s

\pmb {\bf{Hence,}}

Hence,

Hence,

\purple᪣ {\boxed{ \sf{ Rate \: of \: decrease \: of \: slant \: height = \frac{1}{2 \sqrt{3} {\pi}}{cm}^{3} /s }}} ᪣᪣

Rateofdecreaseofslantheight=

2

3

π

1

cm

3

/s

᪣

ᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚᚚ