How to find an exponential function with two points?

1 answer:

6 0

Well u have a basic equation y=a • b^x
a is the initial amount
b is the growth factor
and x is the exponent

the first step is to make a chart with ur two points for example
(0,4) and (2,16)
so ur chart will be
x. | y.
0. | 4
2. | 16
next you find the difference between the x side so 0 to 2 is +2
then find the difference between the y side so 4 to 16 is +12
then put it into a fraction with y over x or y/x so 12/2 then simplified 6/1 or just 6.

6 is the growth factor
and to find a u have to go on the y column and find the first number so
a is 4
x is still x because it's 0 but if it was 2 then it would be x-2 so it can cancel
so the answer would be y=4 • 6^x

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Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues. (assume that

Lilit [14]

Your sequence
-8, 16/3, -32/9, 64/27
is a geometric sequence with first term -8 and common ratio
(16/3)/(-8) = (-32/9)/(16/3) = -2/3

The general term an of a geometric sequence with first term a1 and ratio r is given by
an = a1·r^(n-1)
For your sequence, this is
an = -8·(-2/3)^(n-1)

6 0

3 years ago

Help with this question

Morgarella [4.7K]

Answer:

C) (x-3)(x+3)

Note:

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7 0

3 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$