Answer:
0
Step-by-step explanation:
A= (4-1)^3
simplify A to be 3^3
Which gives us 27
B=(2*3)^2-9
simplify B
first multiply the 2 numbers in paranthesis which gives us 6. raise it to the power of 2 which is 39 and then subtract 9. Gives us 27.
C=15^3*4-12
Simplify the exponent first. 3*4 gives us 12 and 12-12 equals 0. Anything raised to the power of 0=1
If A-B^C is the equation we can write 27-27 raised to the power of 1 which is 0
Answer:
See answer below
Step-by-step explanation:
The possible zeroes are p/q where p is factors of the constant and q is factors of the coefficient of the largest degree.
This means possible zeroes are ±15/4, ±5/4, ±3/4, ±1/4, ±15/2, ±5/2, ±3/2, ±1/2, ±15, ±5, ±3, ±1.
1. 1/9 since you take all the "a"s in the words which is 2/18
2.2/9 same thing as above
3.18/18 since there is no "O" it's a trick question
4.3/9
5.5/6 since you don't count 6
6.1/2 and 1/2 coins are always 1/2 unless you flip twice
7.5/13
8.1/2 because 1,3,5 odd 2,4,6 even
Hope that helps
Any number above 1 gets greater, below 1 smaller (when above 0), while 1 itself remains the same. Negative numbers are more unpredictable.
9 - 6 + 4 - 8/3 ..,
geometric series a(n) = a1r^(n-1)
r = a(n+1)/a(n)
-6/9 = -2/3
4/-6 = -2/3
-8/3/4 = -2/3
so r = -2/3 and a1 = 9
Sn = a1(1-r^n)/(1-r) = 9(1-(-2/3)^n)/(1-(-2/3))
n is infinite Sn = 9/(5/3) = 27/5
