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Kobotan [32]
3 years ago
7

Gas Use

Mathematics
1 answer:
Alisiya [41]3 years ago
7 0
Answer: 10.75
Steps:
After 110miles, it used 2.75gallons
Therefore, the amount of gas used per mile is: 100/2.75= 40 gallon per mile

So after 430miles, the number of gallons used: 430/40 = 10.75 gallons

Hope this helps ʕ•ᴥ•ʔ
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Does the given ordered pair, (-3,5/2) satisfies the given equation. 5x-2y=5 ?
anzhelika [568]

Answer:

No, it does not.

5 0
3 years ago
What is the slope of the line passing through the points (-3,<br> -5) and (-1, - 6)?
Ira Lisetskai [31]

Answer:

-1/2

Step-by-step explanation:

We can find the slope using the slope formula

m = ( y2-y1)/(x2-x1)

   = ( -6 - -5)/( -1 - -3)

   = ( -6+5)/( -1+3)

   =-1/ 2

3 0
3 years ago
Read 2 more answers
The length of a rectangle is 5 ft less than three times the width, and the area of the rectangle is 50 ft². Find the dimensions
Mariulka [41]

Answer

Length = 10 ft

Width = 5 ft

Explanation

Area of the rectangle given = 50 ft²

Let the width of the rectangle be x

So this means the length of the rectangle will be 3x - 5

What to find:

The dimensions of the rectangle.

Step-by-step solution:

Area of a rectangle = length x width

i.e A = L x W

Put A = 50, L = 3x - 5, W = x into the formula.

\begin{gathered} 50=(3x-5)x \\ 50=3x^2-5x \\ 3x^2-5x-50=0 \end{gathered}

The quadratic equation can now be solve using factorization method:

\begin{gathered} 3x^2-5x-50=0 \\ 3x^2-15x+10x-50=0 \\ 3x(x-5)+10(x-5)=0 \\ (3x+10)(x-5)=0 \\ 3x+10=0\text{ }or\text{ }x-5=0 \\ 3x=-10\text{ }or\text{ }x=5 \\ x=-\frac{10}{3}\text{ }or\text{ }x=5 \end{gathered}

Since the dimension can not be negative, hence the value of x will be = 5.

Therefore, the dimensions of the rectangle will be:

\begin{gathered} Length=3x-5=3(5)-5=15-5=10\text{ }ft \\  \\ Width=x=5\text{ }ft \end{gathered}

7 0
1 year ago
In the figure, AB is divided into equal parts. The coordinates of point A are (2, 4), and the coordinates of point Bare (10,6).
olasank [31]

Answer: D = 4, 4.5

E = 5, 4.75

H = 8, 5.5

I = 9, 5.75

Step-by-step explanation:Gang

4 0
3 years ago
The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,
uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right).

We need to solve it and obtain an expression for P(t) in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right).

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that k=0.04 and K=500 and the initial condition P(0)=100.

Notice that this equation is separable, then

\frac{dP}{P(1-P/K)} = kdt.

Now, intagrating in both sides of the equation

\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C.

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}.

So,

\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|.

We have obtained that:

-\ln\left| \frac{K-P}{P}\right| = kt +C

which is equivalent to

\ln\left| \frac{K-P}{P}\right|= -kt -C

Taking exponentials in both hands:

\left| \frac{K-P}{P}\right| = e^{-kt -C}

Hence,

\frac{K-P(t)}{P(t)} = Ae^{-kt}.

The next step is to substitute the given values in the statement of the problem:

\frac{500-P(t)}{P(t)} = Ae^{-0.04t}.

We calculate the value of A using the initial condition P(0)=100, substituting t=0:

\frac{500-100}{100} = A} and A=4.

So,

\frac{500-P(t)}{P(t)} = 4e^{-0.04t}.

Finally, as we want the value of t such that P(t)=200, we substitute this last value into the above equation. Thus,

\frac{500-200}{200} = 4e^{-0.04t}.

This is equivalent to \frac{3}{8} = e^{-0.04t}. Taking logarithms we get \ln\frac{3}{8} = -0.04t. Then,

t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325.

So, the population of rats will be 200 after 25 months.

6 0
4 years ago
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