The value of the equation
Let L and S represent the weights of large and small boxes, respectively. The problem statement gives rise to two equations:
.. 7L +9S = 273
.. 5L +3S = 141
You can solve these equations various ways. Using "elimination", we can multiply the second equation by 3 and subtract the first equation.
.. 3(5L +3S) -(7L +9S) = 3(141) -(273)
.. 8L = 150
.. L = 150/8 = 18.75
Then we can substitute into either equation to find S. Let's use the second one.
.. 5*18.75 +3S = 141
.. S = (141 -93.75)/3 = 15.75
A large box weighs 18.75 kg; a small box weighs 15.75 kg.
Answer:
The 4th one.
Step-by-step explanation:
On all the tables x and y are always increasing, except for the 4th one where the y is decreasing.
Answer:
LHS.= Sin 2x /( 1 + cos2x )
We have , sin 2x = 2 sinx•cosx
And. cos2x = 2cos^2 x - 1
i.e . 1+ cosx 2x = 2cos^2x
Putting the above results in the LHSwe get,
Sin2x/ ( 1+ cos2x ) =2 sinx•cosx/2cos^2x
=sinx / cosx
= Tanx
.•. sin2x/(1 + cos2x)= tanx
Step-by-step explanation:
