You'll have to move the decimal point<span> to the right as many </span>places<span> as there are zeros in the factor,then you'll have to m</span>ove the decimal point<span> one step to the right (10 has one zero),and m</span>ove the decimal point<span> two steps to the right (100 has two zeros).</span>
Answer:
The average is around 75.5
Step-by-step explanation:
Answer:
1/6(1x+1)
Step-by-step explanation:
Step-by-step explanation:
Given P=Rs.2,500,r=4%, N=2
CI=P(1+
100
R
)
2
−P=2,500(1+
100
4
)
2
−2,500=2,500(
100
2
104
2
−1)=
10,000
2,500×816
=Rs.204
SI=
100
P×T×R
=
100
2,500×4×2
=Rs.200
Difference=Rs.4
Guven that the <span>distance
d that a certain particle moves may be calculated from the expression

where a and b are constants; and t is the elapsed time.
Distance is a length and hence the dimension of distance is L.
Now,

and

also will have the dimension of L.
</span><span>Time has a dimension of T.
For

, let the dimension of

be

, then

For

, let the dimension of

be

, then

Therefore, the dimension of

is

while the dimension of

is

.
</span>