All categories

2 years ago

Determine whether each relation is a function if so provide the domain and range

Mathematics

1 answer:

Tems11 [23]2 years ago

4 0

Answers:

Function? Yes
Domain = {1, 5, 10, 11, 14}
Range = {6, 50, 150, 176, 256}

=============================================

Explanation:

x = width

y = area

x is the input while y is the output.

We have a function because each input leads to exactly one output only. We don't have repeated x values.

The domain is the set of possible inputs. So we simply list the set of x values given in the table to get the domain to be {1,5,10,11,14}

The range is a similar story, but we list the possible y outputs getting {6,50,150,176,256}.

The curly braces indicate we have a set of values.

You might be interested in

17. TRAFFIC SIGNS The diagram shows a sign used to warndrivers of a school

MakcuM [25]

each numbered angle is

1. right angle

2. obtuse angle

Road signs, often known as traffic signs, are posted along the side of roadways or above them to inform and direct drivers. Simple wooden or stone milestones were the oldest types of signs. Later, directional signs with arms were developed, such as the fingerposts used in the United Kingdom and their wooden equivalents in Saxony. Since the 1930s, as traffic volumes have increased, many nations have adopted pictorial signs or have similarly standardized and simplified their signs in an effort to overcome linguistic barriers and improve traffic safety. Such pictorial signals are typically based on international norms and use symbols (often silhouettes) in place of words. These signs were initially created in Europe and have since been adopted in varied degrees by the majority of nations.

To know more about traffic signs refer to brainly.com/question/13158950

#SPJ9

The complete question is attached below

7 0

11 months ago

The formula to convert temperature in Celsius to Fahrenheit is F= 9/5C+32. Solve for C<br> C= ?

icang [17]

F = 9/5C + 32

Subtract 32 to both sides:

F - 32 = 9/5C

Divide 9/5 to both sides or multiply by its reciprocal, 5/9:

5/9(F - 32) = C

Simplify:

C = 5/9F - 160/9

6 0

2 years ago

15 POINT 1 QUESTION<br> ITS ALL UP TO YOU!!!!

Anni [7]

First, we can simplify the left side. (6+8x)/2 is equal to 3 + 4x. We can put the equation together at this point.

3 + 4x = 5x

we can subtract 4x from both sides to get our final answer,

3 = x, or x = 3. :)

8 0

2 years ago

A system of equations is given below. For what value(s) of x is f(x)=g(x) ?

NARA [144]

$\large{f(x) = g(x)} \\ \large{ {x}^{2} + 3x = x + 3}$

Since we are solving the quadratic equation because the highest degree in the equation is second. We arrange in the form of ax²+bx+c = 0.

$\large{ {x}^{2} + 3x - x - 3 = 0}$

Combine like terms.

$\large{ {x}^{2} + 2x - 3 = 0}$

Solve the equation by factoring.

$\large{(x + 3)(x - 1) = 0} \\ \large{x = - 3,1}$

Hence the values of x that make f(x) = g(x) are -3 and 1.

Answer

x = -3,1

Let me know if you have any doubts!

3 0

2 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago