Let l, t, b represent the numbers of lions, tigers, bears, respectively.
2l +3t +3b = 156 . . . . . . . 156 meals per day are supplied
l +t = 3b . . . . . . . . . . . . . . there are 3 times as many great cats as bears
l +t +b = 68 . . . . . . . . . . . there are a total of 68 animals
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The last 2 equations tell you
.. 4b = 68
.. b = 17
Subtracting 3 times the last equation from the first gives
.. -l = -48
There are 48 lions, 3 tigers, and 17 bears.
Give me the points thanks
We need to solve the zeroes of the given expression x² - 13x + 30 = 0 and we need to apply zero product property.First, we need to identify the two numbers which will result to -13 when added and it will result to 30 when multiplied. These two numbers are -3 and -10. Then, we can proceed with the solution such as:
x² - 13x + 30 = 0
(x-3) (x- 10) =0
From above, we have already the two zero product:
x-3 = 0
x1 = 3
x-10 =0
x2 =10
The answers are x1 = 3 and x2 = 10.
Out of 28 kids there is one that isnt 12
so 27 kids are 12 years old
1 isnt
concluding that, that 6th grade might be 11
so 1 out of 28
1:28
