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.
Consider this option:
1. formula is: S=0.5*d₁*d₂, where d₁;d₂ - the diagonals of the rhombus.
2. substituting the values into the formula: S=0.5*21*32=336 m².
answer: 336 m².
All your answers look good!
Answer:
Note that
0.16666...is equivalent to 1/6
Step-by-step explanation:
We can use the sum of fractions to write out the decimal:
For this particular question, we can see that
0.16666...is the fraction 1/6, therefore
the given decimal
2.16666...=2+1/6=13/6
Answer:
Darin drove '2m - 20' miles.
Step-by-step explanation:
We are given that Sam drove 'm' no. of miles.
As, it is also given that Kara drove twice as many miles as Sam. Therefore we get that Kara drove '2m' no. of miles.
Now, it is given that Darin drove 20 miles fewer than Kara .i.e. Darin drove '2m - 20' no. of miles.
Hence, in terms of m, Darin drove '2m - 20' miles.
