15 cucumbers to 45 carrots...
so the ratio of cucumbers to carrots is : 15/45 which reduces to 1/3 <==
B. Would be your answer
Using synthetic division, it would turn out to be x+9-9/x+8
Hope I helped!
Answer:
The number of fishes person b initially has is 0
The number of fishes person a initially has is x, where x is any natural number
Step-by-step explanation:
The given parameters are;
let x represent the number of fish person a catches and let y represent the number of fishes that person b catches, we have;
x - x/2 = y + x/2
2·x - x = 2·y + x
2·x = 2·y + 2·x
∴ y = 0
Therefore;
The number of fishes person b initially has = 0
The number of fishes person a initially has = x, where x is any natural number.
Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
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The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
75%
135x100= 13500
13500/180=75
