Answer:
x = 399 new subscribers
Step-by-step explanation:
Let 420 = 100%
x = 95%
420*95 = 100x
x = 420*95/100
x = 399 new subscribers
What is the explicit formula for the sequence?
3, 1, –1, –3, –5, . . .
Answer:
The equation is
.
0.4713 grams would remain in the tumor after 8.5 days.
Step-by-step explanation:
Exponential equation of decay:
The exponential equation for the amount of a substance that decays, after t days, is given by:
![A(t) = A(0)*(1-r)^t](https://tex.z-dn.net/?f=A%28t%29%20%3D%20A%280%29%2A%281-r%29%5Et)
In which A(0) is the initial amount and r is the decay rate, as a decimal.
A tumor is injected with 0.52 grams of Iodine-125.
This means that ![A(0) = 0.52](https://tex.z-dn.net/?f=A%280%29%20%3D%200.52)
After 1 day, the amount of Iodine-125 has decreased by 1.15%.
This means that ![r = 0.0115](https://tex.z-dn.net/?f=r%20%3D%200.0115)
So
![A(t) = A(0)*(1-r)^t](https://tex.z-dn.net/?f=A%28t%29%20%3D%20A%280%29%2A%281-r%29%5Et)
![A(t) = 0.52*(1-0.0115)^t](https://tex.z-dn.net/?f=A%28t%29%20%3D%200.52%2A%281-0.0115%29%5Et)
![A(t) = 0.52*(0.9885)^t](https://tex.z-dn.net/?f=A%28t%29%20%3D%200.52%2A%280.9885%29%5Et)
Then use the formula for A(t) to find the amount of Iodine-125 that would remain in the tumor after 8.5 days.
This is A(8.5).
![A(t) = 0.52*(0.9885)^t](https://tex.z-dn.net/?f=A%28t%29%20%3D%200.52%2A%280.9885%29%5Et)
![A(8.5) = 0.52*(0.9885)^{8.5} = 0.4713](https://tex.z-dn.net/?f=A%288.5%29%20%3D%200.52%2A%280.9885%29%5E%7B8.5%7D%20%3D%200.4713)
0.4713 grams would remain in the tumor after 8.5 days.
Answer:
![\frac{135}{360} \times \pi \times {4}^{2} \\ \frac{45}{120} \times \pi \times 4 \times 4 \\ = 18.84 {ft}^{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B135%7D%7B360%7D%20%20%5Ctimes%20%5Cpi%20%5Ctimes%20%20%7B4%7D%5E%7B2%7D%20%20%5C%5C%20%20%5Cfrac%7B45%7D%7B120%7D%20%20%5Ctimes%20%5Cpi%20%5Ctimes%204%20%5Ctimes%204%20%5C%5C%20%20%3D%2018.84%20%7Bft%7D%5E%7B2%7D%20)
<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
C.
Step-by-step explanation:
You can factor the equation:
![(x-16)(x-2)=0](https://tex.z-dn.net/?f=%28x-16%29%28x-2%29%3D0)
By the zero product property:
![x-16=0\\x=16](https://tex.z-dn.net/?f=x-16%3D0%5C%5Cx%3D16)
![x-2=0\\x=2](https://tex.z-dn.net/?f=x-2%3D0%5C%5Cx%3D2)