The original price is $70.00
The problem states that $59.50 is the discounted price, this means that it is not the 100% price of the aquarium. Since 15% is the discount, the remaining price is equivalent to 85% of the original price, which amount to $59.50.
100% - 15% = 85%
100% - 15% = $59.50
To get original price, we must divide the discounted price by its corresponding percentage.
$59.50 / 85% = original price
$59.50 / 0.85 = $70
To check, we must get the 15% equivalent amount from the original price.
$70 x 15% = $10.50 discount value
$70 - $10.50 = $59.50 discounted price
“Square five times a number” with “a number” being x is the same as saying:
5x²
“three more than the number” is the same as saying:
3 + x
Therefore the sentence “When you square five times a number, you get three more than the number.” is the same as:
5x² = 3 + x
Hope this helps, have a great day! :)
B is greater than or equal to -2.
Answer:
v = 18
Step-by-step explanation:
8/3v = 48
were going to have to get 3 on the other side, so were going to cancel it out and multipliy it because it is a fraction
now we would have 8v = 48*3
48*3=144
now the equation looks like this 8v = 144
to isolate v we can divide 8 on both sides
which brings us to v = 18
High Hopes^^
Barry-
Let
![A(t)](https://tex.z-dn.net/?f=A%28t%29)
denote the amount of salt in the tank at time
![t](https://tex.z-dn.net/?f=t)
. We're given that the tank initially holds
![A(0)=100](https://tex.z-dn.net/?f=A%280%29%3D100)
lbs of salt.
The rate at which salt flows in and out of the tank is given by the relation
![\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dA%7D%7B%5Cmathrm%20dt%7D%3D%5Cunderbrace%7B%5Cdfrac%7B11%5Ctext%7B%20lb%7D%7D%7B1%5Ctext%7B%20gal%7D%7D%5Ctimes%5Cdfrac%7B44%5Ctext%7B%20gal%7D%7D%7B1%5Ctext%7B%20min%7D%7D%7D_%7B%5Ctext%7Brate%20in%7D%7D-%5Cunderbrace%7B%5Cdfrac%7BA%28t%29%7D%7B200%2B%2844-11%29t%7D%5Ctimes%5Cdfrac%7B11%5Ctext%7B%20gal%7D%7D%7B1%5Ctext%7B%20min%7D%7D%7D_%7B%5Ctext%7Brate%20out%7D%7D)
![\implies A'(t)+\dfrac{11}{200+33t}A(t)=484](https://tex.z-dn.net/?f=%5Cimplies%20A%27%28t%29%2B%5Cdfrac%7B11%7D%7B200%2B33t%7DA%28t%29%3D484)
Find the integrating factor:
![\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}](https://tex.z-dn.net/?f=%5Cmu%28t%29%3D%5Cexp%5Cleft%28%5Cdisplaystyle%5Cint%5Cfrac%7B11%7D%7B200%2B33t%7D%5C%2C%5Cmathrm%20dt%5Cright%29%3D%28200%2B33t%29%5E%7B1%2F3%7D)
Distribute
![\mu(t)](https://tex.z-dn.net/?f=%5Cmu%28t%29)
along both sides of the ODE:
![(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}](https://tex.z-dn.net/?f=%28200%2B33t%29%5E%7B1%2F3%7DA%27%28t%29%2B11%28200%2B33t%29%5E%7B-2%2F3%7DA%28t%29%3D484%28200%2B33t%29%5E%7B-1%2F3%7D)
![\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}](https://tex.z-dn.net/?f=%5Cbigg%28%28200%2B33t%29%5E%7B1%2F3%7DA%28t%29%5Cbigg%29%27%3D484%28200%2B33t%29%5E%7B-1%2F3%7D)
![A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt](https://tex.z-dn.net/?f=A%28t%29%3D484%5Cdisplaystyle%5Cint%28200%2B33t%29%5E%7B-1%2F3%7D%5C%2C%5Cmathrm%20dt)
![A(t)=22(200+33t)^{2/3}+C](https://tex.z-dn.net/?f=A%28t%29%3D22%28200%2B33t%29%5E%7B2%2F3%7D%2BC)
Since
![A(0)=100](https://tex.z-dn.net/?f=A%280%29%3D100)
, we get
![100=22(200)^{2/3}+C\implies C\approx-652.39](https://tex.z-dn.net/?f=100%3D22%28200%29%5E%7B2%2F3%7D%2BC%5Cimplies%20C%5Capprox-652.39)
so that the particular solution for
![A(t)](https://tex.z-dn.net/?f=A%28t%29)
is
![A(t)=22(200+33t)^{2/3}-652.39](https://tex.z-dn.net/?f=A%28t%29%3D22%28200%2B33t%29%5E%7B2%2F3%7D-652.39)
The tank becomes full when the volume of solution in the tank at time
![t](https://tex.z-dn.net/?f=t)
is the same as the total volume of the tank:
![200+(44-11)t=500\implies 33t=300\implies t\approx9.09](https://tex.z-dn.net/?f=200%2B%2844-11%29t%3D500%5Cimplies%2033t%3D300%5Cimplies%20t%5Capprox9.09)
at which point the amount of salt in the solution would be