The problem statement gives a relation between the amount removed from one bag and the amount removed from the other. It asks for the amount remaining in each bag. Thus, there are several choices for variables in this problem, some choices resulting in more complicated equations than others.
Let's do it this way: let x represent the amount remaining in bag 1. Then the amount removed from bag 1 is (100-x). The amount remaining in bag 2 is 2x, so the amount removed from that bag is (100-2x). The problem statement tells us the relationship between amounts removed:
... (100 -x) = 3(100 -2x)
... 100 -x -3(100 -2x) = 0 . . . . . . subtract the right side
... 5x -200 = 0 . . . . . . . . . . . . . . eliminate parentheses and collect terms
... x -40 = 0 . . . . . . . . . . . . . . . . .divide by 5
... x = 40 . . . . . . . . . . . . . . . . . . . add 40
- 40 kg is left in the first bag
- 80 kg is left in the second bag
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<u>Check</u>
The amount removed from the first bag is 60 kg. The amount removed from the second is 20 kg. The amount removed from the first bag is 3 times the amount removed from the second bag, as described.
So the equation is n/4-4=14
then you +4 to both sides
n/4-4=14 so now the equation is n/4=18 then you multiply 18 by 4 and get 72
+4 +4
then to check your work
72/4-4=14
so N=72
Hope I helped :)
The answer to the question is c. If you are in multiplying, 8(2)= 16 & 3(3)= 9
Answer:
Step-by-step explanation:
Let the exponential function given in the picture is,
f(x) = a(b)ˣ
Here, a = Initial value
b = Base value
x = Exponent
Since, the graph passes through the points (0, 1) and (1, 3)
For (0, 1),
1 = a(b)⁰
a = 1
For (1, 3),
3 = 1(b)¹
b = 3
Therefore, equation of the given function will be,
f(x) = (3)ˣ
Therefore, Initial value = 1
Base = 3
Asymptote (Horizontal) → y = 0
Vertical asymptote → None
Domain → (-∞, ∞) Or {x | x ∈ R}
Range → (0, ∞) Or {y | y > 0}
Therefore, answers given by Juan are incorrect.
