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2 years ago

How to write the slope intercept equation for points of (6,0) and (0,-5)?

Mathematics

1 answer:

Vadim26 [7]2 years ago

8 0

Answer:

$y=\frac{5}{6} x -5$

Step-by-step explanation:

Hi there!

We are given the points (6,0) and (0, -5), and we want to write the equation of the line containing those points in slope-intercept form

Slope-intercept form can be written as y=mx+b, where m is the slope and b is the y intercept

First, we need to find the slope of the line

The formula for the slope can be written as $\frac{y_2-y_1}{x_2-x_1}$ , where $(x_1, y_1)$ and $(x_2, y_2)$ are points

We have everything we need to find the slope, but let's label the points to avoid confusion

$x_1=6\\y_1=0\\x_2=0\\y_2=-5$

Now substitute these values into the formula

m= $\frac{y_2-y_1}{x_2-x_1}$

m= $\frac{-5-0}{0-6}$

Subtract

m= $\frac{-5}{-6}$

Simplify

m= $\frac{5}{6}$

The slope of the line is 5/6

Let's plug this value into the formula for the equation of the line in slope-intercept form.

We substitute 5/6 for m in y=mx+b:
y= $\frac{5}{6}x+b$

Now we need to find b

As stated before, b is the y intercept, which is the value where the line hits the y axis. The value of x at the y intercept is 0.

One of the points we were given is actually the y intercept; that point is (0, -5); notice how the value of x in this point is 0

The value of b is the value of y in this point, which is -5 in this case.

Substitute -5 as b in the formula.

$y=\frac{5}{6} x -5$

Hope this helps!

See more on this subject here (n.b. the answer uses a different way of solving): brainly.com/question/20891204

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Add the following conceptually and using the standard algorithm 57+58

Snezhnost [94]

The standard algorithm in the US (though this might not be the standard everywhere) is to add the digits in the ones place (carrying the 1 - which represents 10 - if their sum is 10 or more), then to move to the tens place and add the digits there (again carrying the one if their sum is more than 10 - though in this case the one that represents 100) and then we move to the hundreds place and so on.

So, let's do this with the numbers given. 57+58. We would line these up one over the other so that the digits 8 and 7 which are in the ones place are above each other. We add these and get 15 which is one ten and five ones. The 5 gets put down and the 1 gets carried over. Now we focus on the digits in the tens place (here 5 and 5). We add these and to their sum add the 1 we carried. This gives us 11 (but in actuality it means 110). We bring down the leftmost 1 and carry the other. There are no digits in the hundreds place so we being down the 1 we carried earlier.

This gives us an answer of 115.

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3 years ago

Evaluate. -|-2 + 9| i need help

amm1812

Hey there!!

Okay so let's start simplifying this :D

-| -2 + 9| <---- remember that the "vertical bars" are absolute value signs

Absolute value signs will make any values within them into a positive value even if it was negative in the first place

-| 7 | <---- first simplify what is within the absolute value signs

- 7 <----- this is your final answer!!!

I hope this helps you! ^__^

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3 years ago

The circle Ci, intersects the y-axis at two points, one of which is (0.4).

Anuta_ua [19.1K]

Answer:

Part 1) r=5 units (see the explanation)

Part 2) $(x-4)^2+(y-7)^2=25$

Part 3) The center of the circle is (-3,4) and the radius is 4 units

Part 4) see the explanation

Step-by-step explanation:

Part 1)

step 1

Find the center of circle C_1

we know that

The distance between the center and point (0,4) is equal to the radius

The distance between the center and point (4,2) is equal to the radius

Let

(x,y) ----> the coordinates of center of the circle

Remember that

The tangent y=2 (horizontal line) to the circle is perpendicular to the radius of the circle at point (4,2)

That means ----> The segment perpendicular to the tangent is a vertical line x=4

The x-coordinate of the center is x=4

The coordinates of center are (4,y)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Remember

The distance between the center (4,y) and point (0,4) is equal to the radius

The distance between the center (4,y) and point (4,2) is equal to the radius

substitute

$\sqrt{(y-4)^{2}+(4-0)^{2}}=\sqrt{(4-4)^{2}+(y-2)^{2}}$

$\sqrt{(y-4)^{2}+16}=\sqrt{(0)^{2}+(y-2)^{2}}$

squared both sides

$(y-4)^{2}+16=(y-2)^{2}$

solve for y

$y^2-8y+16+16=y^2-4y+4$

$y^2-8y+32=y^2-4y+4\\8y-4y=32-4\\4y=28\\y=7$

The coordinates of the center are (4,7)

step 2

Find the radius of circle C_1

$r=\sqrt{(y-4)^{2}+(4-0)^{2}}$

substitute the value of y

$r=\sqrt{(7-4)^{2}+(4-0)^{2}}$

$r=\sqrt{(3)^{2}+(4)^{2}}$

$r=\sqrt{25}$

$r=5\ units$

Part 2)

Find the equation of the circle C, in standard form.

we know that

The equation of a circle in standard form is

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

where

(h,k) is the center

r is the radius

substitute the given values

$(x-4)^2+(y-7)^2=5^2$

$(x-4)^2+(y-7)^2=25$

Part 3) Another circle C2 has equation x² + y2 + 6x – 8y +9=0

Find the centre and radius of C2

we have

$x^2+y^2+6x-8y+9=0$

Convert to standard form

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

where

(h,k) is the center

r is the radius

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^2+6x)+(y^2-8y)=-9$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^2+6x+9)+(y^2-8y+16)=-9+9+16$

$(x^2+6x+9)+(y^2-8y+16)=16$

Rewrite as perfect squares

$(x+3)^2+(y-4)^2=16$

$(x+3)^2+(y-4)^2=4^2$

therefore

The center of the circle is (-3,4) and the radius is 4 units

Part 4) Show that the circle C2 is a tangent to the x-axis

we know that

If the x-axis is tangent to the circle, then the equation of the tangent is y=0

The radius of the circle must be perpendicular to the tangent

That means ----> The segment perpendicular to the tangent is a vertical line The equation of the vertical line is equal to the x-coordinate of the center

x=-3

The circle C_2, intersects the x-axis at point (-3,0)

Verify

The distance between the center (-3,4) and point (-3,0) must be equal to the radius

Calculate the radius

$r=\sqrt{(0-4)^{2}+(-3+3)^{2}}$