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disa [49]
3 years ago
7

Write the equation in graphing form and the graph the shape. NO LINKS/ASSESSMENT!!

Mathematics
1 answer:
charle [14.2K]3 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

Given

16 + x² + y² - 8x - 6y = 0 ( subtract 16 from both sides )

x² + y² - 8x - 6y = - 16 ( collect x and y terms together )

x² - 8x + y² - 6y = - 16

Use the method of completing the square

add ( half the coefficient of the x/y terms)² to both sides

x² + 2(- 4)x + 16 + y² + 2(- 3)y + 9 = - 16 + 16 + 9

(x - 4)² + (y - 3)² = 9 ← in standard form

with centre (4, 3 ) and r = \sqrt{9} = 3

This is a circle with centre (4, 3 ) and radius 3

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Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form
Alex787 [66]

Answer:

y=-\frac{2}{5}x-\frac{1}{5}

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

  • m is the slope
  • b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

m = -\frac{2}{5}

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-\frac{2}{5}x + b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

  • (-3,1). When x of the line is -3, y of the line must be 1.
  • (2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: y=-\frac{2}{5}x + b. b is what we want, the --\frac{2}{5} is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

  • (-3,1). y = mx + b or 1=-\frac{2}{5} * -3 + b, or solving for b: b = 1-(-\frac{2}{5})(-3).b = -\frac{1}{5}.
  • (2,-1). y = mx + b or -1=-\frac{2}{5} * 2 + b, or solving for b: b = 1-(-\frac{2}{5})(2). b = -\frac{1}{5}.

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points  (-3,1) and (2,-1) is y=-\frac{2}{5}x-\frac{1}{5}

8 0
3 years ago
Which of the following numbers have a value equal to -5
Keith_Richards [23]

We must find which of the following numbers has a value equal to -5:

\frac{-5}{1},\frac{5}{1},\frac{10}{-2},\frac{-5}{-1},-\frac{5}{1},-\frac{-5}{1},-\frac{-5}{-1},-\frac{-10}{2}.

To answer this question, we simplify each number:

\begin{gathered} \frac{-5}{1}=-5 \\ \frac{5}{1}=5 \\ \frac{10}{-2}=\frac{5\cdot2}{-2}=\frac{5}{-1}=-5 \\ -\frac{5}{1}=-5 \\ -\frac{-5}{1}=-(-5)=5 \\ -\frac{-5}{-1}=-\frac{5}{1}=-5 \\ -\frac{-10}{2}=-\frac{-5\cdot2}{2}=-(-5)=5 \end{gathered}

Answer

The numbers that are equal to -5 are:

\begin{gathered} \frac{-5}{1} \\ \frac{10}{-2} \\ -\frac{5}{1} \\ -\frac{-5}{-1} \end{gathered}

4 0
1 year ago
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