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2 years ago

4.0Plss help, will give brianlist

Mathematics

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Helga [31]2 years ago

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Answer:

cat are mammals in some ways

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HELPPP PLEASEEE ILL GIVE B IF CORRECT AND EXPLAIN

timurjin [86]

Answer: 60 degrees

Explanation: If all angles are same, each angle is 60 degrees and adds up to 180.

6 0

3 years ago

Read 2 more answers

Question 1 (1 point) Point M is the midpoint of AB. The coordinates of point A are (-6, 3) and the coordinates of M are (-1, 1).

konstantin123 [22]

Answer:

The answer to your question is: (4, -1)

Step-by-step explanation:

M = midpoint (-1, 1)

A = (-6, 3)

B = (x, y)

$xm = \frac{x1 + x }{2}$

x = 2xm - x1

x = 2(-1) + 6

x = -2 + 6

x = 4

$ym = \frac{y1 + y }{2}$

y = 2ym - y1

y = 2(1) - 3

y = 2 - 3

y = -1

4 0

3 years ago

If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers. Roun

AlekseyPX

Answer:

P = 0.008908

Step-by-step explanation:

The complete question is:

The table below describes the smoking habits of a group of asthma sufferers

Nonsmokers Light Smoker Heavy smoker Total

Men 303 35 37 375

Women 413 31 45 489

Total 716 66 82 864

If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers.

The number of ways in which we can select x subjects from a group of n subject is given by the combination and it is calculated as:

$nCx=\frac{n!}{x!(n-x)!}$

Now, there are 82C2 ways to select subjects that are both heavy smokers. Because we are going to select 2 subjects from a group of 82 heavy smokers. So, it is calculated as:

$82C2=\frac{82!}{2!(82-2)!}=3321$

At the same way, there are 864C2 ways to select 2 different people from the 864 subjects. It is equal to:

$864C2=\frac{864!}{2!(864-2)!}=372816$

Then, the probability P that two different people from the 864 subjects are both heavy smokers is:

$P=\frac{82C2}{864C2}=\frac{3321}{372816}=0.008908$

4 0

3 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

3 years ago

A football field measures 300 feet in length. It can also be expressed as approximately mc019-1.jpg miles. If there are 5,280 fe

WINSTONCH [101]

Can you send some thing i don't get it a bit

7 0

3 years ago