1. Correct answer is a (since when you are not given the n values, it is implied that the first value is 1).
3. Correct answer is a (a(n)=-3+4(1-1)=-3).
7. The pattern is to reduce the previous value by 4, thus the 12th term of the sequence is -24 (a(n)=20 - 4(n-1)=20 - 4(12-1)=-24).
8. Missing term is 26 (32+20=52; 52/2=26).
10. Pattern is 625/1=625; 625/5=125; 125/5=25; 25/5=5, thus. the 10th term is 1/3125 (answer a).
we have
the solution is the interval -------> (3,∞)
therefore
the answer in the attached figure
Answer:
20 units
Step-by-step explanation:
This implies that the square can be divided into four equal L-shaped regions. These regions with respect to transformation forms a square.
Perimeter of the square is 40 units. Since a square has equal length of sides, thus each side of the square is 10 units.
Thus, each L-shape region has dimensions; 8 units, 5 units, 5 units and 2 units.
Perimeter of each L-shape region = the addition of the length of each side of the shape
Perimeter of each L-shape region = 8 + 5 + 5 + 2
= 20 units
Answer:
B, 9.625
Step-by-step explanation:
I know there's a more surefire method of doing this problem but since this is multiple choice, you can use process of elimination to solve it.
Answer:
Step-by-step explanation:
The complete question is
Water flows into a tank according to the rate F(t)= (t+6)/(1+t), and at the same time empties out at the rate E(t)= (ln(t+2))/(t+1), with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest galllon, is in the tank at time t=10 minutes.
Let C(t) be the amount of water in the tank at time t. We now that the rate of change of the tank is given by
![\frac{dC}{dt}=[\tex]rate at which water flows in- rate at which water flows out. Then [tex]\frac{dC}{dt}=\frac{t+6}{t+1}-\frac{\ln(t+2)}{(t+1)}[\tex]so, the desired expression is obtained by integrating with respect to t. This leads us to [tex]C(t) = \int \frac{t+1}{t+1}+ \frac{5}{t+1} - \frac{\ln(t+2)}{(t+1)} dt=t+ 5 \ln (|t+1|)-\int \frac{\ln(t+2)}{(t+1)} dt +C](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdt%7D%3D%5B%5Ctex%5Drate%20at%20which%20water%20flows%20in-%20rate%20at%20which%20water%20flows%20out.%20%3C%2Fp%3E%3Cp%3EThen%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BdC%7D%7Bdt%7D%3D%5Cfrac%7Bt%2B6%7D%7Bt%2B1%7D-%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Eso%2C%20the%20desired%20expression%20is%20obtained%20by%20integrating%20with%20respect%20to%20t.%20%3C%2Fp%3E%3Cp%3EThis%20leads%20us%20to%20%3C%2Fp%3E%3Cp%3E%5Btex%5DC%28t%29%20%20%3D%20%5Cint%20%5Cfrac%7Bt%2B1%7D%7Bt%2B1%7D%2B%20%5Cfrac%7B5%7D%7Bt%2B1%7D%20-%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%3C%2Fp%3E%3Cp%3E%3Dt%2B%205%20%5Cln%20%28%7Ct%2B1%7C%29-%5Cint%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%20%2BC)
Unfortunately, the integral 
cannot be expressed using fundamental functions. So, the problem cannot have an specific function (if you are willing to know the complete answer, the integral of this function uses the polylogarithm function with n=2).
Since you want the exact amount of water at time, you need to give C a value, that is, you need to know an initial condition for the problem. This means, you need to know the amount of water in the tank at time 0
