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melisa1 [442]
3 years ago
8

A line has slope 2/3 and y intercept -2. Which answer is the equation of the line?

Mathematics
2 answers:
Bingel [31]3 years ago
6 0

Answer:

y=2/3x-2

Step-by-step explanation:

All of the given options are in slope intercept form.

Slope intercept form is y=mx+b

This is where m is the slope fo the line and b is the y intercept

That said, we can simply plug the given values into the form.

y=(2/3)x+(-2)

Simplify

y=2/3x-2

Luba_88 [7]3 years ago
6 0

Answer:

A. y= 2/3x-2

Step-by-step explanation:

The slope of a line is written in slope-intercept form, which is:

y= mx+b

where m is the slope and b is the y-intercept.

The line has a slope of 2/3 and a y-intercept of -2. Therefore,

m= 2/3

b= -2

Substitute these values into the slope intercept form.

y= mx+b

y=2/3x+(-2)

Adding a negative number (+-) can be simplified to just a negative(-)

y= 2/3x-2

Therefore, the equation of the line is A. y= 2/3x-2

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If X = 16 feet and Y = 8 feet, what is the area of the figure above ? Use 3.14 for pi .
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Answer:

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Step-by-step explanation:

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3 years ago
Please help me solve this answer it all for me please
mestny [16]

Answer:

The slopes are

m1=\dfrac{2}{5}, m2=-\dfrac{5}{2}

Therefore, the equations are equations of <u>  Perpendicular Lines .</u>

Step-by-step explanation:

Given:

y=\dfrac{2}{5}\times x + 1    ......................Equation ( 1 )

5x+2y=-4\\\\\therefore y = \dfrac{-5}{2}\times x-2   ..............Equation ( 2 )

To Find:

Slope of equation 1 = ?

Slope of equation 2 = ?

Solution:

On comparing with slope point form

y=mx+c

Where,

m = Slope

c = y-intercept

We get

Step 1.

Slope of equation 1 = m1 = \dfrac{2}{5}

Step 2.

Slope of equation 1 = m2 = -\dfrac{5}{2}

Step 3.

Product of Slopes = m1 × m2 = \dfrac{2}{5}\times -\dfrac{5}{2}=-1

Product of Slopes = m1 × m2 = -1

Which is the condition for Perpendicular Lines

The slopes are

m1=\dfrac{2}{5},m2=-\dfrac{5}{2}

Therefore, the equations are equations of <u>  Perpendicular Lines . </u>

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Step-by-step explanation: Khan

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