Answer:
a_n = 2^(n - 1) 3^(3 - n)
Step-by-step explanation:
9,6,4,8/3,…
a1 = 3^2
a2 = 3 * 2
a3 = 2^2
As we can see, the 3 ^x is decreasing and the 2^ y is increasing
We need to play with the exponent in terms of n
Lets look at the exponent for the base of 2
a1 = 3^2 2^0
a2 = 3^1 2^1
a3 = 3^ 0 2^2
an = 3^ 2^(n-1)
I picked n-1 because that is where it starts 0
n = 1 (1-1) =0
n=2 (2-1) =1
n=3 (3-1) =2
Now we need to figure out the exponent for the 3 base
I will pick (3-n)
n =1 (3-1) =2
n =2 (3-2) =1
n=3 (3-3) =0
3x - 4y = 9 and 2x + 3y = 7, which of the following would be the best method?
Let's solve this by using the quadratic formula:
Note that we only use the coefficients so a=12, b=-14, and c=-6.
Plug values in the quadratic equation:
And so by evaluating those values we obtain:
Now we have two answers which are our factors one where we add another where we subtract and so:
First factor:
Second Factor:
And so your factors are
meaning that those are your roots/x-intercepts.
Answer:
The point (3, -3) is on the line.
Step-by-step explanation:
In order to find if it is a solution, input the numbers into the appropriate places in the equation.
y = -2x + 3
-3 = -2(3) + 3
-3 = -6 + 3
-3 = -3
