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2 years ago

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At one point on the trip the miles are recorded as 15,121 when the gas tank is 75% full. Use a ratio equation to determine the m

iles per gallon the sedan averages.

1 answer:

vivado [14]2 years ago

5 0

Answer:

1090miles/gallon approximately

Step-by-step explanation:

On average a Sedan tank can hold 18.5 gallons.

If it's 75% full, it means it is 18.5 x 0.75 = 13.875 gallons full.

Since the number of miles covered was recorded as 15121 with 75% full gas tank, then to determine mile/gallon can be found by

15121/13.875 ≈ 1090miles/gallon

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Plsss help answer the question don’t just get the points and not answer my question Plss I’m struggling

Shkiper50 [21]

Answer:

This is cheating

Step-by-step explanation:

First you follow what your teachers tell you to do and listen

6 0

2 years ago

Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d

Leni [432]

Answers:

(a) $y = \frac{1}{1 - Cx}$ , for any constant C

(b) Solution does not exist

(c) $y = \frac{256}{256 - 15x}$

(d) $y = \frac{64}{64 - 15x}$

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

$x\frac{dy}{dx} = y^2 - y
\\
\\ \indent xdy = \left ( y^2 - y \right )dx
\\
\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}
\\
\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} 
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}$ (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

$\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}
\\
\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} 
\\
\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }$ (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

$1 = (A+B)y - B
\\
\\ \indent \Rightarrow (A+B)y - B = 0y + 1
\\
\\ \indent \Rightarrow \begin{cases}
 A + B = 0
& \text{(3)}\\-B = 1
 & \text{(4)} \end{cases}$

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1

Hence,

$\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}$

So,

$\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} 
\\
\\ \indent \indent \indent \indent = \ln (y-1) - \ln y
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}$

Now, equation (1) becomes

$\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1
\\
\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2
\\
\\ \indent \frac{y-1}{y} = e^{C_1 - C_2}x
\\
\\ \indent \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}
\\
\\ \indent 1 - \frac{1}{y} = Cx
\\
\\ \indent \frac{1}{y} = 1 - Cx
\\
\\ \indent \boxed{y = \frac{1}{1 - Cx}}
$ (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

$

Hence, for any constant C, the following solution will pass thru (0, 1):

$\boxed{y = \frac{1}{1 - Cx}}$

(b) Using equation (5) in problem (a),

$y = \frac{1}{1 - Cx}$ (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that

$y = \frac{1}{1 - Cx}
\\
\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1$

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C.

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 16 = \frac{1}{1 - C(16)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 16C}
\\
\\ \indent 16(1 - 16C) = 1
\\ \indent 16 - 256C = 1
\\ \indent - 256C = -15
\\ \indent \boxed{C = \frac{15}{256}}


$

By replacing this value of C, the general solution becomes

$y = \frac{1}{1 - Cx}
\\
\\ \indent y = \frac{1}{1 - \frac{15}{256}x} 
\\ 
\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}
\\
\\
\\ \indent \boxed{y = \frac{256}{256 - 15x}}



$

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
- Substitute the values of x and y to the general solution.
- Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that

$y = \frac{1}{1 - Cx} 
\\ 
\\ \indent 16 = \frac{1}{1 - C(4)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 4C} 
\\ 
\\ \indent 16(1 - 4C) = 1 
\\ \indent 16 - 64C = 1 
\\ \indent - 64C = -15 
\\ \indent \boxed{C = \frac{15}{64}}$

Now, we replace C using the derived value in the general solution. Then,

$y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}$

5 0

2 years ago

It says that it’s wrong

pochemuha

Answer:

cool

Step-by-step explanation:

what says what is wrong?

3 0

2 years ago

Read 2 more answers

Find the distance between the points with the given coordinates.√ (copy and paste the radical if necessary, no decimals)

Yakvenalex [24]

Answer:

$d=\sqrt{593}$

Step-by-step explanation:

<u>Distance Between two Points in the Plane</u>

Given two points A(x,y) and B(w,z), the distance between them is:

$d=\sqrt{(z-y)^2+(w-x)^2}$

We are given the points (5, 14) and (−3, −9). Substituting into the formula:

$d=\sqrt{(-9-14)^2+(-3-5)^2}$

$d=\sqrt{(-23)^2+(-8)^2}$

$d=\sqrt{529+64}$

$\boxed{d=\sqrt{593}}$

5 0

3 years ago

X+6y=27<br> 7x-3y=9 por metodo de igualacion

GalinKa [24]

Answer:

(3,4)

Step-by-step explanation:

The system of equations is:

x+6y=27

7x-3y=9.

I looked up "metodo de igualacion". It is basically American for doing substitution.

However, the only difference is you are asked to solve both equations for a variable.

The first equation looks easy to solve for x. So I'm going to solve both equations for x.

x+6y=27

Subtract 6y on both sides:

x =-6y+27

7x-3y=9

Add 3y on both sides:

7x =3y+9

Divide both sides by 7:

x =3/7 y +9/7

So both equations are solved for x. You want to find when the x's are the same because you are looking for a common amongst the lines given.

So we have

-6y+27=3/7 y +9/7

I hate the fractions honestly so I'm going to multiply both sides by 7 so they will no longer be for now:

-42y+189=3y + 9

Now add 42y on both sides:

189=45y+9

Subtract 9 on both sides:

180=45y

Divide both sides by 45:

4=y

If 4=y, then y=4.

So now once we have obtain 4 for y, we will use one of the equations given along with it to find x. Just choose one. Choose the easier looking one to you.

I like the x=-6y+27 with y=4.

So replace y with giving you:

x=-6(4)+27

x=-24+27

x=3

So the solution is (x,y)=(3,4).

x=3 and y=4.

8 0

3 years ago