Answer:
<u>Tank 1 : </u>
height = 2x-2
length = 4x
y = l w h
= (4x) (x) (2x-2)
= (4
) (2x-2)
= 8
-8![x^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D)
<u>Tank 2: </u>
height = 6x
length = 2x-1
y = l w h
= (2x-1) (x) (6x)
= (2
-x) (6x)
= 12
-8![x^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D)
I think it is the second option but I am not sure because the picture is very unclear. Do double check with the answers I have worked out.
Hope this helps.
Answer:
![\mathbf{P =\bigg (P_o +\dfrac{ I}{k} \bigg)e^{kt}- \dfrac{I}{k}}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%20%3D%5Cbigg%20%28P_o%20%2B%5Cdfrac%7B%20I%7D%7Bk%7D%20%5Cbigg%29e%5E%7Bkt%7D-%20%5Cdfrac%7BI%7D%7Bk%7D%7D)
Step-by-step explanation:
Given that:
A country population at any given time (t) is:
![\dfrac{dP}{dt}= kP+I](https://tex.z-dn.net/?f=%5Cdfrac%7BdP%7D%7Bdt%7D%3D%20kP%2BI)
where;
P = population at any time t
k = positive constant
I = constant rate of immigration into the country.
Using the method of separation of the variable;
![\dfrac{dP}{kP+1}= dt](https://tex.z-dn.net/?f=%5Cdfrac%7BdP%7D%7BkP%2B1%7D%3D%20dt)
Taking integration on both sides:
![\int \dfrac{dP}{kP+I}= \int \ dt](https://tex.z-dn.net/?f=%5Cint%20%5Cdfrac%7BdP%7D%7BkP%2BI%7D%3D%20%5Cint%20%5C%20dt)
![\dfrac{1}{k} log (kP + I) = t+c_1 \ \ \ here: c_1 = constant \ of \ integration](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bk%7D%20log%20%28kP%20%2B%20I%29%20%3D%20t%2Bc_1%20%5C%20%5C%20%5C%20here%3A%20c_1%20%3D%20constant%20%5C%20of%20%5C%20integration)
![log (kP + I) =k t+kc_1](https://tex.z-dn.net/?f=log%20%28kP%20%2B%20I%29%20%3Dk%20t%2Bkc_1)
By applying the exponential on both sides;
![e^{log (kP + I) }=e^{k t+kc_1 }](https://tex.z-dn.net/?f=e%5E%7Blog%20%28kP%20%2B%20I%29%20%7D%3De%5E%7Bk%20t%2Bkc_1%20%7D)
![KP+I = e^{kt} *e^{kc_1}](https://tex.z-dn.net/?f=KP%2BI%20%3D%20e%5E%7Bkt%7D%20%2Ae%5E%7Bkc_1%7D)
Assume ![e^{kc_1 }= C](https://tex.z-dn.net/?f=e%5E%7Bkc_1%20%7D%3D%20C)
Then:
![kP + I = Ce^{kt}](https://tex.z-dn.net/?f=kP%20%2B%20I%20%3D%20Ce%5E%7Bkt%7D)
![kP = Ce^{kt}-I](https://tex.z-dn.net/?f=kP%20%3D%20Ce%5E%7Bkt%7D-I)
![P =\dfrac{ Ce^{kt}-I}{k} \ \ \---- Let \ that \ be \ equation \ (1)](https://tex.z-dn.net/?f=P%20%3D%5Cdfrac%7B%20Ce%5E%7Bkt%7D-I%7D%7Bk%7D%20%5C%20%5C%20%5C----%20Let%20%5C%20that%20%5C%20be%20%5C%20equation%20%5C%20%281%29)
When time t = 0, The Total population of the country is also ![P_o](https://tex.z-dn.net/?f=P_o)
![P_o = \dfrac{Ce^{0(t)} -I}{k}](https://tex.z-dn.net/?f=P_o%20%3D%20%5Cdfrac%7BCe%5E%7B0%28t%29%7D%20-I%7D%7Bk%7D)
![P_o = \dfrac{Ce^{0} -I}{k}](https://tex.z-dn.net/?f=P_o%20%3D%20%5Cdfrac%7BCe%5E%7B0%7D%20-I%7D%7Bk%7D)
![P_o = \dfrac{C-I}{k}](https://tex.z-dn.net/?f=P_o%20%3D%20%5Cdfrac%7BC-I%7D%7Bk%7D)
C - I = kP₀
C = kP₀ + I
Substituting the value of C back into equation(1), we have:
![P =\dfrac{ (kP_o+1)e^{kt}-I}{k}](https://tex.z-dn.net/?f=P%20%3D%5Cdfrac%7B%20%28kP_o%2B1%29e%5E%7Bkt%7D-I%7D%7Bk%7D)
![P =\dfrac{ (kP_o+1)e^{kt}}{k} - \dfrac{I}{k}](https://tex.z-dn.net/?f=P%20%3D%5Cdfrac%7B%20%28kP_o%2B1%29e%5E%7Bkt%7D%7D%7Bk%7D%20-%20%5Cdfrac%7BI%7D%7Bk%7D)
![\mathbf{P =\bigg (P_o +\dfrac{ I}{k} \bigg)e^{kt}- \dfrac{I}{k}}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%20%3D%5Cbigg%20%28P_o%20%2B%5Cdfrac%7B%20I%7D%7Bk%7D%20%5Cbigg%29e%5E%7Bkt%7D-%20%5Cdfrac%7BI%7D%7Bk%7D%7D)
8(3x−7)=−6(x+7)+4
(8)(3x)+(8)(−7)=(−6)(x)+(−6)(7)+4(Distribute)
24x+−56=−6x+−42+4
24x−56=(−6x)+(−42+4)(Combine Like Terms)
24x−56=−6x+−38
24x−56=−6x−38
Step 2: Add 6x to both sides.
24x−56+6x=−6x−38+6x
30x−56=−38
Step 3: Add 56 to both sides.
30x−56+56=−38+56
30x=18
Step 4: Divide both sides by 30.
30x
/30=
18/30
x=
3/5
Answer:
this is correct on plato/edmentum
Step-by-step explanation:
8)A) 630 square ft 8)B) 3 gallons of paint