There are 5 dimes and 13 nickels
Answer: r = 2/3
Step-by-step explanation:
In a geometric sequence, the consecutive terms differ by a common ratio. The formula for determining the nth term of a geometric progression is expressed as
Tn = ar^(n - 1)
Where
a represents the first term of the sequence.
r represents the common ratio.
n represents the number of terms.
From the information given,
a = 36
The 4th term is 32/3
n = 4
Therefore,
T4 = 32/3 = 36 × r^(4 - 1)
32/3 = 36 × r^3
Dividing both sides if the equation by 36, it becomes
32/3 × 1/36 = r^3
8/27 = r^3
Taking cube root of both sides of the equation, it becomes
(8/27)^1/3 = r^3 × 1/3
r = 2/3
Answer:
Step-by-step explanation:
Given quadratic equation:

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

The above solution is for the quadratic equation of the form:

From the given eqn
a = 1
b = 3
c = 
Now, using the above values in the formula mentioned above:



Now, Rationalizing the above eqn:


Solving the above eqn:

Solving with the help of caculator:

The precise value upto three decimal places comes out to be:

Answer: im pretty sure its 21 mi per gal
Step-by-step explanation:
I think 10. Is 1/20
3/4=15/20
4/5=16/20
So Allison hiked 1/20 more than Patrick