It is true because a rational function is defined as those functions where the variable is placed in the denominator, which must be restricted, because all denominators cannot be equal to zero, other wise it would be undetermined
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.
Answer:
a = 55/6
Step-by-step explanation:
<u>Solving in steps:</u>
- 3/5a = 5 1/2
- 3/5a = 11/2
- a = 11/2 : 3/5
- a = 11/2*5/3
- a = 55/6 or 9 1/6
Answer:
18245
Step-by-step explanation:
We have to use L.C.M,
L.C.M(20,24,32,38)
2|20,24,32,38
2| 10 ,12 ,16 ,19
2| 5 , 6 , 8 , 19
2| 5 , 3 , 4 , 19
2| 5 , 3 , 2 , 19
L.C.M = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 x 19
= 18240
Now for each case remainder is 5,
So the number is 18240+5
=> 18245
the number of elements in the union of the A sets is:5(30)−rAwhere r is the number of repeats.Likewise the number of elements in the B sets is:3n−rB
Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have:150−9x=3n−8x⟺150−x=3n⟺50−x3=n
Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that:150 /10=15is the number of elements in the the A's without repeats counted (same for the Bs as well).So now we have:50−15 /3=n⟺n=45
