Answer:
1
Step-by-step explanation:
(10+x)/3 =(4+x)/1
⇒(10+x)×1=(4+x)×3
10+x=12+3x
⇒-2x-2<0
⇒2x>-2
⇒x>-
2
2
⇒x>-1
this is hard to understand but basically the answer should be 1
Here's the general formula for bacteria growth/decay problems
Af = Ai (e^kt)
where:
Af = Final amount
Ai = Initial amount
k = growth rate constant
<span>t = time
But there's another formula for a doubling problem.
</span>kt = ln(2)
So, Colby (1)
k1A = ln(2) / t
k1A = ln(2) / 2 = 0.34657 per hour.
So, Jaquan (2)
k2A = ln(2) / t
<span>k2A = ln(2) /3 = 0.23105 per hour.
</span>
We need to use the rate of Colby and Jaquan in order to get the final amount in 1 day or 24 hours.
Af1 = 50(e^0.34657(24))
Af1 = 204,800
Af2 = 204,800 = Ai2(e^0.23105(24))
<span>Af2 = 800</span>
Answer:
1/2
Step-by-step explanation:
2, 2 8,5
x1 y1 x2 y2
5-2 3 3/6 would then be simplified to 1/2
------ -
8-2 6
9= h/9
Mutiply both sides by 9
9*9= h/9*9
Cross out 9 and 9 , divide by 9 and then becomes h
9*9= 81
Answer: h= 81