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blondinia [14]
3 years ago
6

Solve the equation x=5/3 pi r ^3 A. B C D?

Mathematics
2 answers:
maks197457 [2]3 years ago
6 0

Answer:

B (3x / ( 5pi)) ^ 1/3 =  r

Step-by-step explanation:

x=5/3 pi r ^3

Multiply each side by 3/5

3/5 x= pi r ^3

Divide each side by pi

3x / ( 5pi) = pi r^3 /pi

3x / ( 5pi) =  r^3

Take the cube root of each side

(3x / ( 5pi)) ^ 1/3 =  r^3 ^ 1/3

(3x / ( 5pi)) ^ 1/3 =  r

Igoryamba3 years ago
4 0
<h2><u>How to solve?</u></h2>

Here, we just have to express r in terms of x when we are provided with the relation between x and r. We have to flip and sent it to the Right hand side.

<h2><u>Solution</u><u>:</u></h2>

We have,

  • \large{ \rm{x =  \frac{5}{3} \pi {r}^{3} }}

Solving it further,

\large{ \longrightarrow{ \rm{x =  \frac{5}{3} \pi {r}^{3} }}}

Taking 5/3 to the other side by dividing it,

\large{ \longrightarrow{ \rm{ \frac{3x}{5}  = \pi {r}^{3} }}}

Now, taking π to the other side by dividing it,

\large{ \longrightarrow{ \rm{  \frac{3x}{5\pi}  =  {r}^{3} }}}

Cube rooting LHS to get <u>r</u>

\large{ \longrightarrow{ \rm{ \sqrt[3]{ \frac{3x}{5\pi} }  = r}}}

Flipping it,

\large{ \longrightarrow{ \rm{r =  \sqrt[3]{ \frac{3x}{5\pi} } }}}

So, the correct option:

\huge{ \boxed{ \bf{ \red{Option \: B}}}}

<u>━━━━━━━━━━━━━━━━━━━━</u>

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Read 2 more answers
rock is thrown upward with a velocity of 27 meters per second from the top of a 23 meter high cliff and it misses the cliff on t
ahrayia [7]

the rock will be at 11 meters from the ground level after 5.92 seconds

Step-by-step explanation:

The motion of the rock is a free-fall motion, since the rock is acted upon the force of gravity only. Therefore, it is a uniformly accelerated motion, so its position at time t is given by the equation:

y=h+ut+\frac{1}{2}at^2

where

h = 23 m is the initial height

u = 27 m/s is the initial velocity, upward

a=g=-9.8 m/s^2 is the acceleration of gravity, downward

t is the time

We want to find the time t at which the position of the rock is

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Substituting and re-arranging the equation, we find

11=23+27t-4.9t^2\\4.9t^2-27t-12=0

This is a second-order equation, which has solutions:

t=\frac{27\pm \sqrt{(-27)^2-4(4.9)(-12)}}{2(4.9)}=\frac{27\pm \sqrt{964.2}}{9.8}

So

t_1 = -0.41 s

t_2=5.92 s

The first solution is negative so we neglect it: therefore, the rock will be at 11 meters from the ground level after 5.92 seconds.

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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