how to make a. lady hot
Answers
Explanation:
Given:-
\sf \: Diameter = 75 \: mDiameter=75m
\sf \: Radius \: = 37.5 \: mRadius=37.5m
\sf \: Outer \: radius = 37.5 + 3.5 = 41 \: mOuterradius=37.5+3.5=41m
Solution:-
{ \underline{ \boxed{ \sf{Outer \: area \: from \: center =\pi {r}^{2} }}}} < /p > < p >
Outerareafromcenter=πr
2
</p><p>
\bullet= \sf \: 3.14 \times {41}^{2}∙=3.14×41
2
\bullet \sf \: = 5278.34 \: {m}^{2}∙=5278.34m
2
{ \underline{ \boxed{ \sf{Inner \: area \: = \: \pi {r}^{2} }}}}
Innerarea=πr
2
\bullet = \sf \: 3.14 \times{37.5}^{2}∙=3.14×37.5
2
\bullet\sf \: = \: 4,415.625 \: m {}^{2}∙=4,415.625m
2
{ \underline{ \boxed{ \sf{Area \: of \: round = Outer \: area \: - \: Inner \:area}}}}
Areaofround=Outerarea−Innerarea
\bullet \sf \: = (5278.34 ) - (4,415.625)∙=(5278.34)−(4,415.625)
\bullet \sf \: = 862.715 \: {m}^{2}∙=862.715m
2
{ \underline{ \boxed{ \sf{Cost \: of \: construction \: = Rs.2.00 \: {m}^{2} }}}}
Costofconstruction=Rs.2.00m
2
{ \underline{ \boxed{ \sf{cost \: of \: construction \: = Rs.2.00 \times 862.715}}}}
costofconstruction=Rs.2.00×862.715
\bullet \sf \: = Rs. \: 1,725.43∙=Rs.1,725.43
The cost of constructing road is Rs.
1,725.43..
Apart from the word, “hot” being used to describe temperature, the term, “Hot girl” at times, may be used to refer to an adult female human that is physically attractive, in such a way that she arouses males just by the males simply looking at them.