All categories

3 years ago

What is the slope of the line whose equation is 5y + 6x - 2 = 0

2 answers:

7 0

Question 251021: What is the slope of the line whose equation is 5y + 6x - 2 = 0?
slope = 6
slope = -6/5
slope = 5/6

koban [17]3 years ago

4 0

All you need to do to find the slope here is put it into y = mx + b format.

5y + 6x - 2 = 0

the goal is to get the y alone, so...

5y = 2 - 6x

divide both sides by 5

y = 2/5 - (6/5)x

then, just to be clearer for y = mx + b ...... y = (-6/5)x + (2/5)

the slope should be (-6/5)

You might be interested in

Please help, question is above in picture

Serhud [2]

-3.5 * 2 = -7
-3.5 * 3n = 10.5n

So there is (-7 - 10.5n) -2.5n

Now combine like terms
10.5n - 2.5n = 8n

ANSWER = -7 + 8n

4 0

3 years ago

Write the linear equation for each graph

Maksim231197 [3]

I’m pretty sure it’s like y=-1/2x

5 0

3 years ago

he gross income of Maurice Vaughn is $785 per week. His deductions are $42.25, FICA tax; $90.33, income tax; 2% state tax; 1% ci

docker41 [41]

Answer:

A. $605.32

Step-by-step explanation:

The state tax, city tax, and retirement fund add to 6% of Vaughn's wages, or ...

... $785 · 6% = $47.10

Then after subtracting all the deductions, Vaughn's take-home pay is ...

... $785 -42.25 -90.33 -47.10 = $605.32 . . . . matches A

3 0

3 years ago

Each friendship bracelet requires

iris [78.8K]

6 feet of string. 2x3=6

5 0

3 years ago

Read 2 more answers

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

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

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

$r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} }$

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

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

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

3 0

3 years ago