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Natali [406]
2 years ago
5

How can you tell if two equations for parallel lines or perpendicular lines. please explain?

Mathematics
2 answers:
kicyunya [14]2 years ago
5 0
To find out if they are parallel we need to see if the gradient is the same, to do this we need to get y in terms of x:
assuming the first equation is x+y+7=0
y=-x-7
and
y=x-3
The gradient is the coefficient of x (the number infront of x)
For equation 1 the gradient is -1, and for number 2 it is 1, therefore they are not parallel.
However to check if they are perpendicular we need to see if their gradients multiply to equal -1.
-1*1=-1 therefore they are perpendicular
sesenic [268]2 years ago
5 0
X + y = 7
<u>x -  y = 3</u>
   <u>2x</u> = <u>10</u>
    2      2
     x = 5
5 + y = 7
<u>-5       -5</u>
     y = 2
(x, y) = (5, 2)
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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&#10;\\&#10;\\ \indent xdy = \left ( y^2 - y \right )dx&#10;\\&#10;\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}&#10;\\&#10;\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} &#10;\\&#10;\\ \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}&#10;\\&#10;\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} &#10;\\&#10;\\ \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&#10;\\&#10;\\ \indent \Rightarrow (A+B)y - B = 0y + 1&#10;\\&#10;\\ \indent \Rightarrow \begin{cases}&#10; A + B = 0&#10;& \text{(3)}\\-B = 1&#10; & \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}

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\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} &#10;\\&#10;\\ \indent \indent \indent \indent = \ln (y-1) - \ln y&#10;\\&#10;\\ \indent  \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}

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\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1&#10;\\&#10;\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2&#10;\\&#10;\\ \indent  \frac{y-1}{y} = e^{C_1 - C_2}x&#10;\\&#10;\\ \indent  \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}&#10;\\&#10;\\ \indent  1 - \frac{1}{y} = Cx&#10;\\&#10;\\ \indent \frac{1}{y} = 1 - Cx&#10;\\&#10;\\ \indent \boxed{y = \frac{1}{1 - Cx}}&#10;       (5)

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

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1&#10;&#10;

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}&#10;\\&#10;\\ \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}&#10;\\&#10;\\ \indent 16 = \frac{1}{1 - C(16)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 16C}&#10;\\&#10;\\ \indent 16(1 - 16C) = 1&#10;\\ \indent 16 - 256C = 1&#10;\\ \indent - 256C = -15&#10;\\ \indent \boxed{C = \frac{15}{256}}&#10;&#10;&#10;

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y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent y = \frac{1}{1 - \frac{15}{256}x} &#10;\\ &#10;\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}&#10;\\&#10;\\&#10;\\ \indent \boxed{y = \frac{256}{256 - 15x}}&#10;&#10;&#10;&#10;

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
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        - 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} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - C(4)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 4C} &#10;\\ &#10;\\ \indent 16(1 - 4C) = 1 &#10;\\ \indent 16 - 64C = 1 &#10;\\ \indent - 64C = -15 &#10;\\ \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
It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other.
zheka24 [161]

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider  some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

P(X=x)=nCx*p^{x}*(1-p)^{(n-x)} with x = 0, 1, 2, 3,…, n

Where

n: number of independent trials

p: success probability  

x: number of successes in n trials

In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)} with x = 0, 1

So,

P(A)=P(X=0)+P(X=1)

P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}

P(A)=0.99^{10}+10*0.01*0.99^{9}

P(A)=0.9957

Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:

3*(1-P(A))*P(A)*P(A) = (1-0.9957)*0.9957*0.9957=0.0127

Finally, we have that the probability of returning exactly one of the three packages is 1.27%

3 0
2 years ago
The time in seconds that it takes an object to fall d feet can be found using the
Serga [27]

Answer:

0.3

Step-by-step explanation

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2 years ago
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