Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). & Problem 2: If a piece of real estate purchased for $75,000 in 1998 appreciates at the rate of 6% per year, then its value t years after the purchase will be f(t)=75,000(1.06^t). According to this model, by how much will the value of this piece of property increase between the years 2005 and 2008. & Problem 3: The amount A in an account after t years of an initial principle P invested at an annual rate r compounded continuously is given by A=Pe^rt where r is expressed as a decimal. What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously?
2006-11-22 12:22:41 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4).
You find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)
The horizontal asymptote is y = c, so c = -2. That means b must be [you figure it out]
Now it is y = 4a^x - 2. This must be true when x=1 and y=4.
So 4 = [you figure it out; answer deleted due to complaint from below]
Prob. 3. What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously?
Just plug in the values.
A = 500 e^(.05 x 10)
[answer deleted]
2006-11-22 12:56:08 · answer #1 · answered by hayharbr 7 · 0⤊ 0⤋
U will find the same answer by the other Guy he is my Dam bloody Friend and he stole the answer and Pasted.
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4).
You find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)
The horizontal asymptote is y = c, so c = -2. That means b must be [you figure it out]
Now it is y = 4a^x - 2. This must be true when x=1 and y=4.
So 4 = [you figure it out; answer deleted due to complaint from below]
Prob. 3. What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously?
Just plug in the values.
A = 500 e^(.05 x 10)
2006-11-23 01:51:00 · answer #2 · answered by integral_op 3 · 0⤊ 0⤋
1. y intercept 2 means when x = 0, y = 2, so sub that in (remembering a^0 = 1) and you have one equation connecting b and c.
You know how the graph of y = a^x approaches the x axis out on the left? because very big negative values of x give very tiny values of y. Now in the equation we have here, very big negative values of x make the first term very close to zero, and so the horizontal asymptote is y = c. So now you know the value of c. Sub it in the first equation we got, and you find b as well.
Finally sub x = 1 and y = 4, and get another equation which you cna solve to find a. If you can't, and someone else doesn't come in and tell you the answer (as usual), email h_chalker@yahoo.com.au
2006-11-22 13:00:42 · answer #3 · answered by Hy 7 · 0⤊ 0⤋
Problem 2 is the easiest of all. Simply run your numbers to find the amount built up in 2005 (114538.61) and the amount in 2008 (142382.39) and subtract the smaller from the larger.
This leaves you with an amount of 27,833.79 (give or take a few pennies due to rounding).
2006-11-22 12:51:10 · answer #4 · answered by saberhilt 4 · 0⤊ 0⤋
-7= 3x-8 even as putting -8 to the different aspect of equation, the signal will substitute, so that's -7+8= 3x which ability a million=3x for this reason x= a million/3 and don't be disturbed about the order, it doesn't remember! there's a rule in math about the order! you could actual examine it by substituting a million/3 into the equation i.e -7= 3*a million/3-8 -7=-7!!!
2016-11-29 09:27:45 · answer #5 · answered by kuebler 4 · 0⤊ 0⤋