Perimeter: 2w+2l
We know the width and the perimeter, which is 18 and 76, so let's plug it in.
76=2(18)+2l
76=36+2l isolate the variable x!
40=2l
20=l
every length is 20 inches.
Total earned Friday = $1445
Total earned Saturday = $1624
Earnings Sunday - $xxx
Average over 3 day period = $1500
Therefor total over 3 day = $1500 x 3 = $4500
So earnigs friday + Saturday = $1445 + $1624 = $3069
$4500 - $3069 = $1431 earned on Sunday
Answer:
The Time period required for decay of Iodine-125 to half of its value is 60 days .
Step-by-step explanation:
Given as :
The initial quantity of iodine-125 = 0.4 gram
The rate of decay = 1.15 %
Let The time period for decay = x day
The finial quantity after decay = half of initial quantity
I.e The finial quantity after decay = 0.2 gram
Now ,
The final quantity after decay = initial quantity × ![(1-\dfrac{\textrm rate}{100})^{\textrm Time}](https://tex.z-dn.net/?f=%281-%5Cdfrac%7B%5Ctextrm%20rate%7D%7B100%7D%29%5E%7B%5Ctextrm%20Time%7D)
Or, 0.2 gm = 0.4 gm × ![(1-\dfrac{\textrm 1.15}{100})^{\textrm x}](https://tex.z-dn.net/?f=%281-%5Cdfrac%7B%5Ctextrm%201.15%7D%7B100%7D%29%5E%7B%5Ctextrm%20x%7D)
or,
= ![(0.9885)^{x}](https://tex.z-dn.net/?f=%280.9885%29%5E%7Bx%7D)
Or, 0.5 = ![(0.9885)^{x}](https://tex.z-dn.net/?f=%280.9885%29%5E%7Bx%7D)
Or,
= 0.9885
Taking log both side
log (
) = Log 0.9885
or,
× log 0.5 = - 0.0050233
or,
× ( - 0.301029 ) = - 0.0050233
or, x =
∴ x = 59.92 ≈ 60 days
Hence The Time period required for decay of Iodine-125 to half of its value is 60 days . Answer
Answer: 70
Step-by-step explanation: 6300/90=70 Brainliest please?