(a) I have attached the graph.
(b) The function g(x) is given by g(x) = f(2x + 2)/3. You can work this out algebraically. You can also work this out by using the graph of the function.
(c) We can obtain the graph of y = g(x) from the graph of y = f(x) through the following transformations:
* stretch horizontally by a factor of 2
* stretch vertically by a factor of 3
* shift downwards by 2 units
Answer:
the answer is h
Step-by-step explanation:
The sum clearly diverges. This is indisputable. The point of the claim above, that
is to demonstrate that a sum of infinitely many terms can be manipulated in a variety of ways to end up with a contradictory result. It's an artifact of trying to do computations with an infinite number of terms.
The mathematician Srinivasa Ramanujan famously demonstrated the above as follows: Suppose the series converges to some constant, call it
. Then
Now, recall the geometric power series
which holds for any
. It has derivative
Taking
, we end up with
and so
But as mentioned above, neither power series converges unless
. What Ramanujan did was to consider the sum
as a limit of the power series evaluated at
:
then arrived at the conclusion that
.
But again, let's emphasize that this result is patently wrong, and only serves to demonstrate that one can't manipulate a sum of infinitely many terms like one would a sum of a finite number of terms.
The matchings are as follows:
1) 2=3 , 4 =5 (Given)
2) 1 =3 (vertical angles are equal)
3) 1 =2 (substitution)
4) VR=VR (Reflexive)
5) Triangle VSR congruent to Triangle VTR (ASA)
6) RS=RT CPCTE
The matchings are done above. The triangles are congruent as per ASA congruency as two angles and corresponding side is equal.
since 2 =3 and 1 = 3 we get 1=2
By CPCTE we have RS =RT
Answer:
35
Step-by-step explanation:
Here we see 5 black keys for every 7 white keys.
So the ratio is 5:7
If we need 49 white keys, find the amount we scale the original ratio by:
49/7 = 7
So we are scaling by a factor of 7.
The number of black keys would be 5 * the scale of 7. = 35
So there should be 35 black keys.
