Answer:
16
Step-by-step explanation:
Not an expertise on infinite sums but the most straightforward explanation is that infinity isn't a number.
Let's see if there are anything we missed:
∞
Σ 2^n=1+2+4+8+16+...
n=0
We multiply (2-1) on both sides:
∞
(2-1) Σ 2^n=(2-1)1+2+4+8+16+...
n=0
And we expand;
∞
Σ 2^n=(2+4+8+16+32+...)-(1+2+4+8+16+...)
n=0
But now, imagine that the expression 1+2+4+8+16+... have the last term of 2^n, where n is infinity, then the expression of 2+4+8+16+32+... must have the last term of 2(2^n), then if we cancel out the term, we are still missing one more term to write:
∞
Σ 2^n=-1+2(2^n)
n=0
If n is infinity, then 2^n must also be infinity. So technically, this goes back to infinity.
Although we set a finite term for both expressions, the further we list the terms, they will sooner or later approach infinity.
Yep, this shows how weird the infinity sign is.
The probability would be out of 100,
in this case if it refers to the data from 2009 it would be 39:61
Answer: HUHHHHHHHHHHHHHHH
Step-by-step explanation:
Answer:
First odd number = 13
Second odd number = 15
Step-by-step explanation:
Let
First odd number = 2x + 1
Second odd number = 2x + 3
According to given conditions:
32 + 2x + 1= 3 (2x + 3)
33 + 2x = 6x + 9
Taking terms with x to left and constants to right
2x - 6x = 9 - 33 (sign of transferred terms will be changed)
-4x = -24
Dividing both sides by -4
x = -24/-4
x =6
So,
First odd number = 2x + 1 = 2*6 + 1 = 12 + 1 =13
Second odd number = 2x + 3= 2*6 +3= 12 + 3 = 15
I hope it will help you!
