Can someone please please help with these questions?

Does this table represent an exponential function x 1 2 3 4 y -1 -8 -27 -64

k0ka [10]

$\bf \begin{array}{ccll}
x&y\\
\cline{1-2}
1&\stackrel{-1^3}{-1}\\\\
2&\stackrel{-2^3}{-8}\\\\
3&\stackrel{-3^3}{-27}\\\\
4&\stackrel{-4^3}{-64}\\\\
x&-x^3
\end{array}~\hspace{5em}\boxed{y=-x^3}~~\checkmark$

4 0

2 years ago

Find four distinct complex numbers (which are neither purely imaginary nor purely real) such that each has an absolute value of

Luda [366]

Answer:

0.5 + 2.985i
1 + 2.828i
1.5 + 2.598i
2 + 2.236i

Explanation:

Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.

Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.

The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:

$|C| = \sqrt{a^2+b^2}$

Then, the work consists in finding pairs (a,b) for which:

$\sqrt{a^2+b^2}=3$

You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:

$\sqrt{a^2+b^2}=3\\ \\ a^2+b^2=3^2\\ \\ a^2=9-b^2\\ \\ a=\sqrt{9-b^2}$

I will use b =0.5, b = 1, b = 1.5, b = 2

$b=0.5;a=\sqrt{9-0.5^2}=2.958\\ \\b=1;a=\sqrt{9-1^2}=2.828\\ \\b=1.5;a=\sqrt{9-1.5^2}=2.598\\ \\b=2;a=\sqrt{9-2^2}=2.236$

Then, four distinct complex numbers that have an absolute value of 3 are:

0.5 + 2.985i
1 + 2.828i
1.5 + 2.598i
2 + 2.236i

4 0

3 years ago

Mrs. Shultz went to New York to do her Christmas shopping. She took a cab to go into the city and they charged her a flat fee of

Orlov [11]

Answer:

12 miles

Step-by-step explanation:

30-6=24

24/2=12 miles

3 0

2 years ago

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha

8_murik_8 [283]

To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
where
$A$ is the final pupolation
$P$ is the initial population
$e$ is the Euler's constant
$r$ is the decay rate
$t$ is the time in years

A. We know for our problem that the initial population is 1,250, so $P=1250$ ; we also know that after a year the population is 1000, so $A=1000$ and $t=1$ . Lets replace those values in our formula to find $r$ :
$A=Pe^{rt}$
$1000=1250e^{r}$
$e^{r}= \frac{1000}{1250}$
$e^{r}= \frac{4}{5}$
$ln(e^{r})=ln( \frac{4}{5} )$
$r=ln( \frac{4}{5} )$
$r=-02231$

Now that we have $r$ , we can write a function to model this scenario:
$A(t)=1250e^{-0.2231t}$ .

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C.
- The function is decreasing
- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:
$A(0)=1250e^{(-0.2231)(0)$
$A_{0}=1250e^{0}$
$A_{0}=1250$
- Over the interval [0,10], the function will have a minimum at t=10:
$A(10)=1250e^{(-0.2231)(10)$
$A_{10}=134.28$

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: $m= \frac{A(0)-A(10)}{10-0}$
where
$m$ is the rate of change
$A(10)$ is the function evaluated at 10
$A(0)$ is the function evaluated at 0
We know from previous calculations that $A(10)=134.28$ and $A(0)=1250$ , so lets replace those values in our formula to find $m$ :
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
$m=-111.572$
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

7 0

2 years ago

Please answer this question.

kobusy [5.1K]

Answer:

its i think the first one and the last one

Step-by-step explanation:

sorry if its wrong but i think it is right

5 0

2 years ago