All categories

2 years ago

Help me please asap i do not understand this.

Mathematics

1 answer:

Tpy6a [65]2 years ago

5 0

The range of the given graph is equal to (2, 7).

<h3>What is a range?</h3>

A range can be defined as the set of all real numbers that connects with the elements of a domain. This ultimately implies that, a range refers to the set of all possible output numerical values, which are shown on the y-axis of a graph.

<h3>How to identify the range and domain of this graph?</h3>

The vertical extent of a graph represents all range values and they are always read and written from smaller to larger numerical values, and from the bottom of the graph to the top.

Similarly, the horizontal extent of a graph represents all domain values and they are also read and written from smaller to larger numerical values, and from the left of the graph to the right.

By critically observing the graph of y(x) shown, we can infer and logically deduce the following:

Domain = (-6, 6)

Range = (2, 7).

Read more on Range here: brainly.com/question/17003159

#SPJ1

You might be interested in

Find an integer x such that 0<=x<527 and x^37===3 mod 527

Greeley [361]

Since $527=17\times31$ , we have that

$x^{37}\equiv3\mod{527}\implies\begin{cases}x^{37}\equiv3\mod{17}\\x^{37}\equiv3\mod{31}\end{cases}$

By Fermat's little theorem, and the fact that $37=2(17)+3=1(31)+6$ , we know that

$x^{37}\equiv(x^2)^{17}x^3\equiv x^5\mod{17}$
$x^{37}\equiv(x^1)^{31}x^6\equiv x^7\mod{31}$

so we have

$\begin{cases}x^5\equiv3\mod{17}\\x^7\equiv3\mod{31}\end{cases}$

Consider the first case. By Fermat's little theorem, we know that

$x^{17}\equiv x^{16}x\equiv x\mod{17}$

so if we were to raise $x^5$ to the $n$ th power such that

$(x^5)^n\equiv x^{5n}\equiv x\mod{17}$

we would need to choose $n$ such that $5n\equiv1\mod{16}$ (because $16+1\equiv1\mod{16}$ ). We can find such an $n$ by applying the Euclidean algorithm:

$16=3(5)+1$
$\implies1=16-3(5)$
$\implies16-3(5)\equiv-3(5)\equiv1\mod{16}$

which makes $-3\equiv13\mod{16}$ the inverse of 5 modulo 16, and so $n=13$ .

Now,

$x^5\equiv3\mod{17}$
$\implies (x^5)^{13}\equiv x^{65}\equiv x\equiv3^{13}\equiv(3^4)^2\times3^4\times3^1\mod{17}$

$3^1\equiv3\mod{17}$
$3^4\equiv81\equiv4(17)+13\equiv13\equiv-4\mod{17}$
$3^8\equiv(3^4)^2\equiv(-4)^2\mod{17}$
$\implies3^{13}\equiv(-4)^2\times(-4)\times3\equiv(-1)\times(-4)\times3\equiv12\mod{17}$

Similarly, we can look for $m$ such that $7m\equiv1\mod{30}$ . Apply the Euclidean algorithm:

$30=4(7)+2$
$7=3(2)+1$
$\implies1=7-3(2)=7-3(30-4(7))=13(7)-3(30)$
$\implies13(7)-3(30)\equiv13(7)equiv1\mod{30}$

so that $m=13$ is also the inverse of 7 modulo 30.

And similarly,

$x^7\equiv3\mod{31}[/ex] [tex]\implies (x^7)^{13}\equiv3^{13}\mod{31}$

Decomposing the power of 3 in a similar fashion, we have

$3^{13}\equiv(3^3)^4\times3\mod{31}$

$3\equiv3\mod{31}$
$3^3\equiv27\equiv-4\mod{31}$
$\implies3^{13}\equiv(-4)^4\times3\equiv256\times3\equiv(8(31)+8)\times3\equiv24\mod{31}$

So we have two linear congruences,

$\begin{cases}x\equiv12\mod{17}\\x\equiv24\mod{31}\end{cases}$

and because $\mathrm{gcd}\,(17,31)=1$ , we can use the Chinese remainder theorem to solve for $x$ .

Suppose $x=31+17$ . Then modulo 17, we have

$x\equiv31\equiv14\mod{17}$

but we want to obtain $x\equiv12\mod{17}$ . So let's assume $x=31y+17$ , so that modulo 17 this reduces to

$x\equiv31y+17\equiv14y\equiv1\mod{17}$

Using the Euclidean algorithm:

$17=1(14)+3$
$14=4(3)+2$
$3=1(2)+1$
$\implies1=3-2=5(3)-14=5(17)-6(14)$
$\implies-6(14)\equiv11(14)\equiv1\mod{17}$

we find that $y=11$ is the inverse of 14 modulo 17, and so multiplying by 12, we guarantee that we are left with 12 modulo 17:

$x\equiv31(11)(12)+17\equiv12\mod{17}$

To satisfy the second condition that $x\equiv24\mod{31}$ , taking $x$ modulo 31 gives

$x\equiv31(11)(12)+17\equiv17\mod{31}$

To get this remainder to be 24, we first multiply by the inverse of 17 modulo 31, then multiply by 24. So let's find $z$ such that $17z\equiv1\mod{31}$ . Euclidean algorithm:

$31=1(17)+14$
$17=1(14)+3$

and so on - we've already done this. So $z=11$ is the inverse of 17 modulo 31. Now, we take

$x\equiv31(11)(12)+17(11)(24)\equiv24\mod{31}$

as required. This means the congruence $x^{37}\equiv3\mod{527}$ is satisfied by

$x=31(11)(12)+17(11)(24)=8580$

We want $0\le x$ , so just subtract as many multples of 527 from 8580 until this occurs.

$8580=16(527)+148\implies x=148$

3 0

4 years ago

Which of the following units are units of volume? Select all that apply.

oksano4ka [1.4K]

Answer:

we have these we ml and l

Step-by-step explanation:

Here, we want to select the units of volume ;

1. mg

This is a measure of mass and in fact it is called milligrams

2. mm

This is called millimeters and in fact is a measure of distance

3. cm

This is centimeters and it is a measure of distance too

4. ml

Milliliters, this is a measure of volume

5. m

meters;

this is a measure of distance

6. g

grammes

This is a measure of mass

7. l

This is liters

It is actually a measure of volume

8. km

this is kilometers, it is a measure of distance

7 0

3 years ago

What is the greatest common factor of 24x^4 and 80x^5

Ronch [10]

I hope this helps you

24x^4 80x^5 ÷ x^4 ✔

24 80 x ÷ 8 ✔

3 10 x

GCF=8.x^4

7 0

3 years ago

-2(3)+8-10=<br> I don't know how to simplify this

Tanzania [10]

-8 hope it helps!!!!!!!!!!!!!!!!!!!

5 0

3 years ago

Sucesión II: 3, 7, 11, 15, 19, 23 ... ¿Cual es la expresión algebraica que la genera? * x=4n-2 x= 4n-1 x=3n-2

LenaWriter [7]

Responder:

4n-1

Explicación paso a paso:

Dada la secuencia

3, 7, 11, 15, 19, 23 ..

El enésimo término si la secuencia se expresa como;

Tn = una + (n-1) d

a es el primer término = 3

n es el número de términos

d es la diferencia común = 7-3 = 11-7

d = 4

Sustituir

Tn = 3+ (n-1) (4)

Tn = 3 + 4n-4

Tn = 4n-1

Por lo tanto, el enésimo término de la secuencia es 4n-1

8 0

3 years ago