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3 years ago

For the function {(2, 4), (3, 5), (4, 6), (5, 7)}, what is the domain?

Mathematics

1 answer:

sammy [17]3 years ago

6 0

Answer: Domain:

{

}

Step-by-step explanation:

The domain is the set of all the values of

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An airplane travels 6903 kilometers against the wind in 9 hours and 8613 kilometers with the wind in the same amount of time. Wh

Rzqust [24]

Answer:

The rate of the plane is 862 miles per hour

Step-by-step explanation:

An airplane travels 6903 kilometers against the wind in 9 hours.

Speed = distance / time

Speed of the airplane while travelling against the wind is

6903/9 = 767 miles per hour

The airplane travelled 8613 kilometers with the wind in the same amount of time. This means that the speed while travelling with the wind will be

8613/9 = 957 miles per hour

Let x = speed of the airplane

Let y = speed of the wind

While travelling against the wind,

x - y = 767 - - - - - -- 1

While travelling with the wind,

x + y = 957 - - - - - - -2

Subtracting equation 2 from equation 1,

-y - y = 767 - 957

-2y = -190

y = -190/-2

y = 95 miles per hour

x = 957 - y

x = 957 - 95

x = 862 miles per hour

4 0

3 years ago

15-x=2(x+3) I need this solved and show me how you solved it

Sunny_sXe [5.5K]

15 - x = 2(x + 3)
15 - x = 2x + 6
15 = 3x + 6
9 = 3x
3 = x
x = 3

4 0

3 years ago

A salesman sells a car for $37,000 and earns 8.5% commission. How much commission does he earn?

nevsk [136]

Answer:

4,352 dollars

Step-by-step explanation:

6 0

2 years ago

If the figure is reflected over the x-axis, what are the coordinates of point R?

12345 [234]

Answer:

R' (3,-3)

Step-by-step explanation:

4 0

3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago