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Explain how to modify the graphs of f(x) and g(x) to graph the solution set to the following system of inequalities. How can the

Norma-Jean [14]

The graphs of f(x) and g(x) are transformed function from the function y = x^2

The set of inequalities do not have a solution

<h3>How to modify the graphs</h3>

From the graph, we have:

$f(x) = x^2 -3$ and $g(x) = -x^2 +2$

To derive y < x^2 - 3, we simply change the equality sign in the function f(x) to less than.

To derive y > x^2 + 2, we perform the following transformation on the function g(x)

Shift the function g(x) down by 2 units
Reflect across the x-axis
Shift the function g(x) down by 3 units
Change the equality sign in the function g(x) to greater than

<h3>How to identify the solution set</h3>

The inequalities of the graphs become

y < x^2 - 3 and y > x^2 + 2

From the graph of the above inequalities (see attachment), we can see that the curves of the inequalities do not intersect.

Hence, the set of inequalities do not have a solution

Read more about inequalities at:

brainly.com/question/25275758

7 0

3 years ago

URGENT PLEASE HELP<br> x/240 = 82/100 what is X?

Zielflug [23.3K]

x/240 = 82/100

240 : 100 = 2.4

so 240 is divided with 2,4 giving 100, this will work with 82 too. So, 82 × 2.4 = 196,8

Answer : x = 196,8

prove :

196,8/240 = 82/100 = 41/50

see the picture

7 0

3 years ago

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Can some one help me with the whole paper I don’t get it

valina [46]

Answer:

I think-

1. 11

2. 0.7

3. 67.89

4. 13

Step-by-step explanation:

I'm not 100% sure tho

5 0

4 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

3 years ago

PLEASE HELP the information is below

liq [111]

Answer:

yellow: x=6

red: y=4

blue: y=-2

Step-by-step explanation:

3 0

3 years ago

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