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Vinvika [58]
3 years ago
10

Which inequality is represented by the graph y>-2/3x+1 y<-2/3x+1 y<-3/2x+1 y>-3/2x+1

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer:

the 3rd one is the graph answer

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Find the equation of the line perpendicular to y=3/2x-9 and passes through the point (6,2)​
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y=-2/3x-2

Step-by-step explanation:

slope= -2/3

y=-2/3x+b

enter the given points

2=4+b

b=-2

y=-2/3x-2

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What does x = in this equation x+3+2x=x+5
Ede4ka [16]

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X would be equal to 1

Step-by-step explanation:


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Find the derivative of the fucntion. Simplify where possible.<br>y = tan^-1(x²)​
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Write an equation of the line that passes through 7,10 and is perpendicular to the line y=1/2-9
kow [346]

An equation of the line that passes through (7,10) and is perpendicular to the line y=1/2x-9 is y=-2x+24

Step-by-step explanation:

We need to write an equation of the line that passes through 7,10 and is perpendicular to the line y=1/2x-9

<u>Note:</u> Considering the line y=1/2x-9

For finding an equation of line we need

  • Slope (m)
  • y-intercept (b)

The general equation of slope intercept form used is:

y=mx+b

The equation of line given is: y=1/2x-9

The new line is perpendicular to the given line So, the slope of new line would be m = -1/m

So, Slope of given line m = 1/2

Slope of required line = -1/m = -2

So, m= -2

Now, for finding y-intercept (b) we need slope m = -2 and the point given (7,10)

Putting values of m, x=7 and y = 10 in the equation of slope intercept form

y=mx+b\\10=-2(7)+b\\10=-14+b\\b=10+14\\b=24

So, Equation of required line having slope m = -2 and y-intercept b = 24

y=mx+b\\y=-2x+24

An equation of the line that passes through (7,10) and is perpendicular to the line y=1/2x-9 is y=-2x+24

Keywords: Equation of line using Slope  

Learn more about Equation of line using Slope at:

  • brainly.com/question/4819659
  • brainly.com/question/4097107
  • brainly.com/question/1979240

#learnwithBrainly

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3 years ago
Name this set of numbers which -2/5 belongs ?
Damm [24]
That belongs in the rational #s
5 0
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