Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Answer:
3x + 13y
Explanation:
6x + 8y − (3x − 5y)
Distribute the Negative Sign
6x + 8y + −1(3x − 5y)
6x + 8y + −1(3x) + −1(−5y)
6x + 8y + −3x + 5y
Combine Like Terms
6x + 8y + −3x + 5y
(6x + −3x) + (8y + 5y)
= 3x + 13y
The Real Answer is B: Leasing a car for a certain period.
Hope this helps!
Answer:
19
Step-by-step explanation:
3.920(1.018)ˣ = 5.500
- First we can divide both sides by 3.920 to get: (1.108)ˣ = 1.403
- Then we can take the natural log of both sides to get: ln[(1.108)ˣ] = ln(1.403)
- We can then use logarithmic rules to move the x exponent to the front: xln(1.108) = ln(1.403)
- Then we can divide both sides by ln(1.108) to get:
- Then using a calculator we get x = 18.98 ≈19
