Can someone help me please

The entire graph of the function g is shown in the figure below.

marusya05 [52]

The domain of the function is [-4, 4) and the range of the function is [-5, 2)

<h3>How to determine the domain and the range of the function?</h3>

<u>The domain</u>

As a general rule, it should be noted that the domain of a function is the set of input values or independent values the function can take.

This means that the domain is the set of x values

From the graph, we have the following intervals on the x-axis

x = -4 (closed circle)

x =4 (open circle)

This means that the domain of the function is [-4, 4)

<u>The range</u>

As a general rule, it should be noted that the range of a function is the set of output values or dependent values the function can produce.

This means that the range is the set of y values

From the graph, we have the following intervals on the y-axis

y = -5 (closed circle)

y = 2 (open circle)

This means that the range of the function is [-5, 2)

Read more about domain and range at:

brainly.com/question/10197594

#SPJ1

7 0

2 years ago

A soccer field is a rectangle 48 meters wide and 55 meters long. The coach asks players to run from one corner to the corner dia

Sindrei [870]

Answer:

Distance = 73 m

Step-by-step explanation:

Given that,

The length of a rectangle = 48 m

The width of a rectangle = 55 m

We need to find the diagonal distance. We can use the Pythagoras theorem to find it.

$d^2=48^2+55^2\\\\d=\sqrt{5329}\\\\d=73\ m$

So, the required distance is equal to 73 m.

5 0

2 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

__

Using these formulas on problem (d), we get ...

d)

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

__

And for problem (f), we get ...

f)

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

2 years ago

Will give Brainlest Answer please answer i need help

Rashid [163]

Answer:

C. {-1, 5, 8}

Step-by-step explanation:

Use each of the domain values in the function to see what the corresponding range value is.

f(-1) = 5 -3(-1) = 8

f(0) = 5 -3(0) = 5

f(2) = 5 -3(2) = -1

The range is the set of numbers {-1, 5, 8}.

_____

<em>Additional comment</em>

The values in a set are generally listed lowest to highest. The coefficient of x in the equation for f(x) is negative, meaning the lowest range value will correspond to the highest domain value. If you start by finding f(2) = -1, you immediately eliminate all answer choices except B and C.

Those choices differ only in the middle value, so you can tell which is correct by evaluating f(x) for the middle domain value: f(0) = 5. Only one answer choice has both -1 and 5 in the set.

(There are two answers here: how you work the problem, and how you game a multiple choice question.)

4 0

2 years ago

In a quiz contest, 4 marks were awarded for a correct answer and -2 marks were given for an incorrect answer. If James gave 5 in

Zina [86]

9514 1404 393

Answer:

10

Step-by-step explanation:

For c correct answers and 5 incorrect answers, James's score is ...

4c -2(5) = 30

4c = 40 . . . . . . add 10

c = 10 . . . . . . . . divide by 4

James gave 10 correct answers.

4 0

2 years ago

Read 2 more answers