.To really see what happens visually, go to fooplot.com and enter the
base function y = e^x to look at its graph (In the textedit field below
the label "Function y(x)", just enter "e^x" and hit the "Enter" key on
your keyboard ... you don't need to enter the "y="). Then type in each
of the example functions provided below to see what happens based upon
the constant introduced.
y = e^x is an exponential
function that increases gradually from (-âž, 0) until it reaches the
point (0,1). Thereafter, it very sharply increases on (0,âž).
.....If you introduce a constant A > 1,
.....y = Ae^x will increase gradually from (-âž, 0) until it reaches the point (0,A).
.....Thereafter, it will rise faster than the base function y = e^x
.....Example. y = 5e^x
.....If your constant A is less than 1 but still positive: 0 < A < 1,
.....y = Ae^x will increase gradually from (-âž, 0) until it reaches the point (0,A).
.....Thereafter, it will have a slower rise on (0,âž) than the base function y = e^x.
.....Example. y = (1/5) e^x
.....If your constant A = -1,
.....y = Ae^x will look like y = Ae^x when A=1 except that the graph will be flipped over the x-axis.
.....Instead of increasing gradually on (-âž, 0), it will decrease gradually on (-âž, 0)
.....until it passes through the point (0, -1) and then veer downward on (0,âž)
.....Example. y = -e^x
.....If your constant A is between -1 and 0: -1 < A < 0,
.....y = Ae^x will be flipped over the x-axis and decrease gradually from(-âž, 0)
.....until it reaches the point (0,A).
.....Thereafter, it will have a slower decrease on (0,âž)
.....Example. y = (-1/5) e^x
.....If your constant A is less than 0: A < -1,
.....y = Ae^x will be reflected (flipped) over the x-axis and decrease gradually from (-âž, 0)
.....until it reaches the point (0,A). Thereafter, it will sharply veer downward.
.....Example. y = -5 e^x
y = e^x + A will translate the base function y = e^x by A units. If A
> 0, it will essentially shift the graph upward by A. If A < 0, it
will shift the graph downward by A.
The main flaw of this proof is that it did not take into consideration the fact a 4 cent postage must be made with a 4 cent stamp and a 3 cent postage must be made with a 3 cent stamp but only looked at allocating the stamps based on the total amount of postage
Step-by-step explanation:
From the question we are told that
The proof is
Every postage stamp of three cents or more can be formed using just 3 cent and 4 cent stamps.
Basic Step : We can form postage of 3 cents with a single 3 cent stamp and we can form postage of 4 cents using a single four cents tamp
Inductive Step: Assume that we can form postage of j cents for all non negative integers j with j <= k using just 3 cent and 4-cent stamps. We can then form postage of k+1 cents by replacing one 3 cent stamp with a 4-cent or two 4-cent stamps with three 3-cent stamps
FLAW IDENTIFICATION
Looking the inductive step, the statement ''by replacing one 3 cent stamp with a 4-cent or two 4-cent stamps with three 3-cent stamps '' means that it is possible to use a 3-cent stamp for postage that requires a 4 -cent stamp and this is not correct according the statement in the basic step
Generally the main flaw of this proof is that it did not take into consideration the fact a 4 cent postage must be made with a 4 cent stamp and a 3 cent postage must be made with a 3 cent stamp but only looked at allocating the stamps based on the total amount of postage
Step-by-step explanation: Let’s say your bank gives you a rate of 2.10 %,
If you purchase a CD $125,000 for a 3 year term. Your APY would be $8,041.52 for 3 years. So at the end of your CD term, you should have a balance of $133,041.52. Hope this gives you an idea.