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:
The coordinates of T' will be T'(13,-1)
Step-by-step explanation:
step 1
Find the rule of the translation
take any point of rectangle JKLM (pre-image)
The coordinate of point k(-3,2)
The coordinate of point k'(4,7)
so
The rule of the transformation
k(-3,2) -----> k'(4,7)
is equal to
(x,y) ------> (x+7,y+5)
That means ----> the translation is 7 units at right and 5 units up
step 2
Apply the rule of the translation at the coordinate T of trapezoid STUV
we have
T(6,-6)
(x,y) ------> (x+7,y+5)
T(6,-6) ------> T'(6+7,-6+5)
T(6,-6) ------> T'(13,-1)
therefore
The coordinates of T' will be T'(13,-1)
Hello there again. I have your answer: Solve for x by simplifying both sides of the equation, then isolating the variable. x = −8
Answer:
4 rows
Step-by-step explanation:
GCF of 12 and 16 is4
Answer: The required inverse transform of the given function is
Step-by-step explanation: We are given to find the inverse Laplace transform, f(t), of the following function :
We have the following Laplace formula :
Therefore, we get
Thus, the required inverse transform of the given function is
