There is no solution since the lines are parallel. (See attached graph)
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Hope this helps!</em></u></h2>
So what my brother does is he places all the points on the graph and lightly go over it joining the points and then go over normally
Solve for x by simplifying both sides of the equation, then isolating the variable.
x = 3y − 11
Answer:
Step-by-step explanation:
a_1 = 15
a_2 = a_1 - 3
a_2 = 15 - 3
a_2 = 12
a_3 = a_2 - 3
a_3 = 12 - 3
a_3 = 9
a_4 = a_3 - 3
a_4 = 9 - 3
a_4 = 6
a_5 = 3
a_6 = 0 I'm leaving these last two to expand
a_n = a1 - (n - 1)*d
a_n = 15 - (n - 1)*3
a_n = 15 - 3n + 3
a_n = 18 - 3n
Example
a_6 = 18 - 3*6
a_6 = 0
Problem B
t(1) = 108
t(1 + 1) = 1/3 * 108
t(2) = 36
t(3) = 1/3 * t2
t(3) = 1/3 * 36
t(3) = 12
t(4) =1/3 (t(3))
t(4) = 1/3 * 12
t(4) = 4
t(5) = t4 / 3
t(5) = 4 / 3
t(5) = 1.3333333
So the explicit definition is
t(n) = 108 (1/3)^(n - 1) You could simplify this a little bit by realizing that 108 is made of three 3s.
t(n) = 4 * 3^3 * (1/3)^(n - 1)
t(n) = 4 * (1/3) ^ (n - 4)
Example
t(5) = 108 (1/3)^4
t(5) = 108(1/81)
t(5) = 1.3333333
And using the simplified formula, you get.
t(5) = 4 * (1/3)^1
t(5) = 1.333333 which is the same thing as the original result.
Answer:
I would need to be able to see the expressions available.
Step-by-step explanation:
