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2 years ago

What is the domain and range for 2x+3y=18? What is the domain and range of 3x-4y>16?

Mathematics

2 answers:

kvv77 [185]2 years ago

5 0

3x - 4y = 9 and 2x + 3y = 7, which of the following would be the best method?

Virty [35]2 years ago

5 0

The domain of a function is the complete set of possible values of the independent variable.

The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.

2x+3y=18

For this equation, domain and range both would be all real numbers.

3x-4y>16

Here domain and range both will be all real numbers.

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Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work

valina [46]

Answer:

Step-by-step explanation:

Question

Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.

perimeter of a triangle = AB+AC+BC

Using the distance formula

AB = sqrt(-2-5)²+(2-2)²

AB = sqrt(-7)²

AB =sqrt(49)

AB =7

BC = sqrt(-2+2)²+(2-5)²

BC = sqrt(0+3²)

BC =sqrt(9)

BC =3

AC= sqrt(-2-5)²+(2-5)²

AC= sqrt(-7)²+3²

AC =sqrt(49+9)

AC =sqrt58

Perimeter = 10+sqrt58

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2 years ago

Let y represent the total cost of publishing a book (in dollars). Let x represent the number of the copies of the book printed.

Oduvanchick [21]

Im sorry idk this answer

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3 years ago

Finding the vaule of an angle

Sergio039 [100]

(3x-5)+(x+1)=180
x=46
3(46)-5=133
m

6 0

3 years ago

Read 2 more answers

Use the Algebra Tiles to construct the algebraic expression for this situation:

Zigmanuir [339]

That’s a hard one use Brainly to find the answer

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3 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

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10 months ago