Question

$\sqrt[3]{8000} + ( \frac{8}{10} ) {3}$
a.20.008
b.20.08
c.20.8
d.208​

Answer 1

Answer:

Step-by-step explanation:

s = ut + 1/2 at2  

is the second equation of motion.  

It is also called position-time relation.  

Using this formula we can calculate the displacement when the values for time, acceleration and initial velocity are known.

 

s - displacement  

u - initial velocity  

t - time  

a - acceleration.  

 

In case of free fall, this equation can be written as

h = ut + 1/2 gt2  

where, h - height from which the object falls on the ground from rest OR the maximum height up to which the object reaches when thrown upward

g - acceleration due to gravity ( value of g is negative when the object is thrown upward and is positive when the object falls to the ground from height)

when the object falls to the ground from height, its initial velocity (u) is zero.  

when the ball reaches the maximum height when thrown in upward direction, its final velocity (v) is zero.

Answer 2

Answer:

$\sqrt[3]{800} = \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5} \\ = 2 \sqrt[3]{100} \\ = 9.283</p><p>$

Step-by-step explanation:

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