52. Rahzel wants to determine how much gasoline they had every month
=> He used 78 1/3 gallons of gas monthly
=> his wife used 41 3/8 gallons of gas last month
How much is to total of gas that they both used.
First let’s convert this fraction to decimal
=> 78 1/3 = 78.33
=> 41 3/8 = 41.38
Now, let’s start adding.
=> 78.33 + 41.38 = 119.71 this is already rounded to the nearest hundredths.
S = d/t
st = d
t = d/s
The time going is t1.
The time returning is t2.
The total time is 4 hours, so we have t1 + t2 = 4
The speed of the current is c.
The speed going is 9 + c.
The speed returning is 9 - c.
t1 = 16/(9 + c)
t2 = 16/(9 - c)
t1 + t2 = 16/(9 + c) + 16/(9 - c)
4 = 16/(9 - c) + 16/(9 + c)
1 = 4/(9 - c) + 4/(9 + c)
(9 + c)(9 - c) = 4(9 - c) + 4(9 + c)
81 - c^2 = 36 - 4c + 36 + 4c
81 - c^2 = 72
c^2 = 9
c^2 - 9 = 0
(c + 3)(c - 3) = 0
c + 3 = 0 or c - 3 = 0
c = -3 or c = 3
We discard the negative answer, and we get c = 3.
The speed of the current is 3 mph.
65-9=56
56 divided by 8 is 7
So they will need 7 tents
Answer:
We want a polynomial of smallest degree with rational coefficients with zeros in 
, 
and -3. The last root gives us the factor (x+3). Hence, our polynomial is
where 
is a polynomial with rational coefficients and roots and . The root gives us a factor 
, but in order to obtain rational coefficients we must consider the factor 
.
An analogue idea works with . For convenience write 
. This gives the factor 
. Hence,
Notice that 
. So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is
Step-by-step explanation:
We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type 
will introduce in the expression, we need to multiply by its conjugate 
. Hence, we will obtain that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.
