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GrogVix [38]
3 years ago
12

What is The equation of a vertical line passing through the point (-5,-1)

Mathematics
2 answers:
Lana71 [14]3 years ago
8 0

Answer:

The answer is x=-5

Step-by-step explanation:

In order to determine the equation, we need to know about vertical line equations.

In general, a line equation has the form:

y=mx+n

Where "y" is the dependent variable, "x" is the independent variable, "m" is the slope of the line and "n" is the value where the line cuts the "y" axis.

A vertical line, x only takes one value. Thus, the equation for a vertical line is x = a, where a is the value that x takes.

So for the coordinate (-5,-1), the vertical line has to pass through the "x" value, that it is -5, therefore the equation of a vertical line passing through the point (-5,-1) is:

x=-5

mihalych1998 [28]3 years ago
3 0
The equation of a vertical line passing through the point (-5,-1) is x = -5

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Answer:

2 1/14

Step-by-step explanation:

First, we will equalize the denominators of the fractions.

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Then we can solve this equation.

2 2/14 - 1/14 = 2 1/14

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Answer:

The second option (see attached image)

Step-by-step explanation:

You are looking for a box diagram that represents 9 units, and from those, clearly marked sections that contain 3/2 = 1.5 units.The idea is to count how many 1.5 units you have in 9 units.

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2 years ago
What is the equation of x
masya89 [10]

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x=-48/5

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3 years ago
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Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

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

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

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

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

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

3 0
3 years ago
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