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##A=(s^2sqrt(3))/4## ##A=(256sqrt(3))/4## ##A=64sqrt(3)/4##

The area is approximately ##86.6 cm^2##. As this hexagon is regular you can divide it into ##6## triangles. Please note that all those triangles are isosceles. […]

##6/8 = 3/4##

##sec^2 x – tan^2 x = 1## Note that: ##sin^2 x + cos^2 x = 1## Hence: ##cos^2 x = 1 – sin^2 x## and we […]

##20 ft## The volume of a cylinder can be found through the formula: ##V=pir^2h## Substitute the givens to solve for ##h##. ##720pi=pi(6)^2h## ##720pi=36pih## Divide both sides […]

one half the product of the diagonals

The half of ##135^o## is ##67.5^o## Using the Half Angle Formula: ##sin (theta/2) = +-sqrt((1-costheta)/2)## ##sin (135^o/2) = +-sqrt((1-cos135^o)/2)## ##cos135^o = cos(90^o +45^o)= cos90^ocos45^o-sin90^oc0s45^o = cancel(0(sqr2/2))-(1)(sqrt2/2) […]

Solve ##cos^2 x = 3/4## Ans: ##+- pi/6## and ##+- (5pi)/6## ##cos^2 x = 3/4## –> ##cos x = +- sqrt3/2## ##a. cos x = sqrt3/2 […]

Since ##(AM)/(MB)=16/8!=20/16=(AL)/(LC)## lines ##ML## and ##BC## are NOT parallel. Since ##(BK)/(KC)=15/12=20/16=(AL)/(LC)## lines ##LK## and ##AB## are parallel. Let’s first formulate triangle proportionality theorem and a converse […]