Answer: After 10 more years i.e. after 30 years, the population of the town will be 100626.57 approx.
Step-by-step explanation:
Since we have given that
Initial population 20 years ago = 10,000
Rate of growth = 8%
Number of years = 10+20 = 30 years
According to question, we will use "Compound interest":
![Amount=P(1+\frac{r}{100})^n\\\\Amount=10000(1+\frac{8}{100})^{30}\\\\Amount=10000(1+0.08)^{30}\\\\Amount=10000(1.08)^{30}\\\\Amount=100626.57](https://tex.z-dn.net/?f=Amount%3DP%281%2B%5Cfrac%7Br%7D%7B100%7D%29%5En%5C%5C%5C%5CAmount%3D10000%281%2B%5Cfrac%7B8%7D%7B100%7D%29%5E%7B30%7D%5C%5C%5C%5CAmount%3D10000%281%2B0.08%29%5E%7B30%7D%5C%5C%5C%5CAmount%3D10000%281.08%29%5E%7B30%7D%5C%5C%5C%5CAmount%3D100626.57)
Hence, After 10 more years i.e. after 30 years, the population of the town will be 100626.57 approx.
Answer:
Step-by-step explanation:
they want you to remember your algebra... :P
mulitply both sides by 5
5*
= -7 * 5 this is fair.. b/c you have done the same thing to both sides of the equal sign... the equal sign remains true
now cancel the 5 s on the left
x = -35 and voila you got it
<span>2x + y = 5
6x + 3y = 25
Multiply both sides of the first equation by 3.
6x + 3y = 15
Now subtract it from the second equation.
0 = 10
Since 0 = 10 is a false statement, there is no solution.
Answer: b. inconsistent, none
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Hello,
I am going to remember:
y'+3y=0==>y=C*e^(-3t)
y'=C'*e^(-3t)-3C*e^(-3t)
y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9
==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k
Well, I'm going to guess the "x" is a multiplication sign. To solve this you have your equation (7/8 x c) and you replace c with 8. Now your equation is 7/8 x 8. To solve the fraction equation you do 7/8 x 8/1 or 7/ So the answer to this equation is 7. Please add as brainiest :)