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4 years ago

15

Find the distance between these two points (-1,0), (8,6)

1 answer:

julia-pushkina [17]4 years ago

8 0

The distance formula is the radical (x2-x1)2 + (y2-y1)^2. So for the x value, you subtract (-1) from 8. But be careful. It’s not 8-1. IT’S 8-(-1). Which x value, 9 should be squared. Which it is 81. The y value, you subtract 0 from 6. Which you should have 6 as your y value. Then, again you have to squared it. So, 6^2 is 36. Now that you have both of your x and y value, you have to subtract them again. In order to subtract, you subtract the y value from the x value. In other words, 81-36. The answer should be 45. But 45 is irrational. Because it can’t be rational under the radical form. So to simplified it, you have to find a number that is rational enough to get out from the inside of the radical form. The only number that would work is 9 and 5. So rad 9 is rational because it could escape from the inside. Which it’s 3 once it’s out. But, 5 is irrational and can’t go out. As a result, the answer is 3 rad 5. Or the 3 is outside of the radical form and 5 is the inside.

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3 years ago

Using the following equation, find the center and radius: x2 −2x + y2 − 6y = 26 (5 points)

Lisa [10]

Answer:

Center: (1,3)

Radius: 6

Step-by-step explanation:

Hi there!

$x^2-2x + y^2 - 6y = 26$

Typically, the equation of a circle would be in the form $(x-h)^2+(y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius.

To get the given equation $x^2-2x + y^2 - 6y = 26$ into this form, we must complete the square for both x and y.

<u>1) Complete the square for x</u>

Let's take a look at this part of the equation:

$x^2-2x$

To complete the square, we must add to the expression the square of half of 2. That would be 1² = 1:

$x^2-2x+1$

Great! Now, let's add this to our original equation:

$x^2-2x+1+y^2-6y = 26$

We cannot randomly add a 1 to just one side, so we must do the same to the right side of the equation:

$x^2-2x+1+y^2-6y = 26+1\\x^2-2x+1+y^2-6y = 27$

Complete the square:

$(x-1)^2+y^2-6y = 27$

<u>2) Complete the square for y</u>

Let's take a look at this part of the equation $(x-1)^2+y^2-6y = 27$ :

$y^2-6y$

To complete the square, we must add to the expression the square of half of 6. That would be 3² = 9:

$y^2-6y+9$

Great! Now, back to our original equation:

$(x-1)^2+y^2-6y+9= 27$

Remember to add 9 on the other side as well:

$(x-1)^2+y^2-6y+9= 27+9\\(x-1)^2+y^2-6y+9= 36$

Complete the square:

$(x-1)^2+(y-3)^2= 36$

<u>3) Determine the center and the radius</u>

$(x-1)^2+(y-3)^2= 36$

$(x-h)^2+(y-k)^2=r^2$

Now, we can see that (1,3) is in the place of (h,k). 36 is also in the place of r², making 6 the radius.

I hope this helps!

3 0

3 years ago

Read 2 more answers

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3 years ago

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