1.208 because 30.2 devided by 25 would be the amount of ounces in each serving
In order to use the remainder theorem, you need to have some idea what to divide by. The rational root theorem tells you rational roots will be from the list derived from the factors of the constant term, {±1, ±5}. When we compare coefficients of odd power terms to those of even power terms, we find their sums are equal, which means -1 is a root and (x +1) is a factor.
Dividing that from the cubic, we get a quotient of x² +6x +5 (and a remainder of zero). We recognize that 6 is the sum of the factors 1 and 5 of the constant term 5, so the factorization is
... = (x +1)(x +1)(x +5)
... = (x +1)²(x +5)
_____
The product of factors (x +a)(x +b) will be x² + (a+b)x + ab. That is, the factorization can be found by looking for factors of the constant term (ab) that add to give the coefficient of the linear term (a+b). The numbers found can be put directly into the binomial factors to make (x+a)(x+b).
When we have 1·5 = 5 and 1+5 = 6, we know the factorization of x²+6x+5 is (x+1)(x+5).
Answer:
9,14,19
Step-by-step explanation:
You seem to be giving the expression 2n+3n+4.
Now simplify
An = 5n+4
Fill in n
5(1) +4
5(2) +4
5(3)+4
After solving and simplifying you see 9, 14 and 19 are the first 3 terms.
Answer:
The total volume of the water in the tank after 20 minutes = 1220 gallons
Step-by-step explanation:
Rate of water pumped into the tank r (t) = 30 (1 - 
)
Initial volume of water in the tank = 800 gallons
The water in the tank after 20 minutes = Initial volume of water in the tank + Volume of water being pumped in the tank
= 
Where a = 0 , b = 20
Put the value of r (t) in above equation we get
= 
= ![30 [ t + \frac{e^{-0.16t} }{0.16} ]](https://tex.z-dn.net/?f=30%20%5B%20t%20%2B%20%5Cfrac%7Be%5E%7B-0.16t%7D%20%7D%7B0.16%7D%20%5D)
= 
= 420 gallon
Now, total volume in the tank
Therefore the total volume of the water in the tank after 20 minutes = 1220 gallons
Answer: -5100
<u>Step-by-step explanation:</u>
![\sum^4_1[100(-4)^{n-1}]\qquad \rightarrow \qquad a_1=100\ \text{and r = -4}\\\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\\\S_4=\dfrac{100(1-(-4)^4)}{1-(-4)}\\\\\\.\quad=\dfrac{100(1-256)}{1+4}\\\\\\.\quad=\dfrac{100(-255)}{5}\\\\.\quad=20(-255)\\\\.\quad=-5100\\](https://tex.z-dn.net/?f=%5Csum%5E4_1%5B100%28-4%29%5E%7Bn-1%7D%5D%5Cqquad%20%5Crightarrow%20%5Cqquad%20a_1%3D100%5C%20%5Ctext%7Band%20r%20%3D%20-4%7D%5C%5C%5C%5C%5C%5CS_n%3D%5Cdfrac%7Ba_1%281-r%5En%29%7D%7B1-r%7D%5C%5C%5C%5C%5C%5C%5C%5CS_4%3D%5Cdfrac%7B100%281-%28-4%29%5E4%29%7D%7B1-%28-4%29%7D%5C%5C%5C%5C%5C%5C.%5Cquad%3D%5Cdfrac%7B100%281-256%29%7D%7B1%2B4%7D%5C%5C%5C%5C%5C%5C.%5Cquad%3D%5Cdfrac%7B100%28-255%29%7D%7B5%7D%5C%5C%5C%5C.%5Cquad%3D20%28-255%29%5C%5C%5C%5C.%5Cquad%3D-5100%5C%5C)
