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VARVARA [1.3K]
3 years ago
8

Y=-3+6and y=1/3x-8 determine if the equation are parallel, perpendicular or neither

Mathematics
1 answer:
Gelneren [198K]3 years ago
7 0

Slope intercept form: y = mx + b

m = slope and b = y-intercept

The equation is: y = -3x + 6

The m value for this is -3, so the line has a slope of -3. If y = 1/3x - 8 was parallel to y = -3x + 6 then they would have the same slope, which they do not. So they aren't parallel.

The find the slope of a perpendicular line, take the negative reciprocal of the slope of y = -3x + 6. The negative reciprocal of -3 is 1/3.  y = 1/3x - 8 has a slope of 1/3, so they are perpendicular to each other.

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Find the center of a circle whose end points of a diameter are A (-5,6) and B (3,4)
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<u>Answer:</u>

(-1, 5)

<u>Step-by-step explanation:</u>

The center of a circle is the midpoint of the diameter.

So we have to calculate the midpoint of the diameter, using this formula:

midpoint = (  \frac{x_{2 }+x_{1}  }{2}  , \frac{y_{2} + y_{1} }{2} )

               =  (\frac{3 + (-5)}{2}  ,  \frac{4 + 6}{2}  )

               = (-1, 5)            

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malfutka [58]

Answer:

  eight more than a fifth of a number

Step-by-step explanation:

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 One of the first procedures used in solving equations has an application in our everyday world. Suppose that we place a -kilogram box on one side of a seesaw and a -kilogram stone on the other side. If the center of the box is the same distance from the balance point as the center of the stone, we would expect the seesaw to balance. The box and the stone do not look the same, but they have the same value in weight. If we add a -kilogram lead weight to the center of weight of each object at the same time, the seesaw should still balance. The results are equal.

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The Addition Principle

If the same number is added to both sides of an equation, the results on each side are equal in value.

We can restate it in symbols this way.

For real numbers a, b, c if a=b thenat+tc=b+ec

Here is an example.

If

, then

Since we added the same amount  to both sides, each side has an equal value.

We can use the addition principle to solve an equation.

EXAMPLE 1 Solve for .   

  Use the addition principle to add   to both sides.

  Simplify.

  The value of  is .

 We have just found the solution of the equation. The solution is a value for the variable that makes the equation true. We then say that the value, , in our example, satisfies the equation. We can easily verify that  is a solution by substituting this value in the original equation. This step is called checking the solution.

Check.    =

         ≟

         =   ✔

 When the same value appears on both sides of the equals sign, we call the equation an identity. Because the two sides of the equation in our check have the same value, we know that the original equation has been correctly solved. We have found the solution.

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 It does not matter which side of the equation contains the variable. The  term may be on the right or left. In the next example the x term will be on the right.

EXAMPLE 2 Solve for .   

  Add  to both sides, since  is the additive inverse of  . This will eliminate the   on the right and isolate .

  Simplify.

  The value of  is .

Check.    =

         ≟   Replace  by .

         =   ✔   Simplify. It checks. The solution is .

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EXAMPLE 3 Solve for .   

  Simplify by adding.

  Add the value   to both sides, since   is the additive inverse of .

  Simplify. The value of  is .

Check.    =

         ≟    Replace  by  in the original equation.

           ✔    It checks.

 In Example 3 we added   to each side. You could subtract  from each side and get the same result. In earlier lesson we discussed how subtracting a  is the same as adding a negative . Do you see why?

 We can determine if a value is the solution to an equation by following the same steps used to check an answer. Substitute the value to be tested for the variable in the original equation. We will obtain an identity if the value is the solution.

EXAMPLE 4 Is  the solution to the equation  ? If it is not, find the solution.

We substitute  for  in the equation and see if we obtain an i

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