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NikAS [45]
3 years ago
13

Two lines, A and B, are represented by equations given below: Line A: y = x − 4 Line B: y = 3x + 4 Which of the following shows

the solution to the system of equations and explains why? (−3, −5), because the point satisfies one of the equations (−3, −5), because the point lies between the two axes (−4, −8), because the point satisfies both equations (−4, −8), because the point does not lie on any axis
Mathematics
2 answers:
Ede4ka [16]3 years ago
5 0
(-4,-8) because it satisfies both equations
notsponge [240]3 years ago
3 0

Answer:

The solution is (-4,-8)because the point satisfies both equations.

Step-by-step explanation:

Line A: y = x − 4

Line B: y = 3x + 4

Substitute the value of y from Line A in Line B

x-4=3x+4

-4-4=2x

-8=2x

-4=x

Substitute the value of x in Line B to get value of y

y=3x+4

y=3(-4)+4

y=-12+4

y=-8

Thus the solution is (-4,-8)because the point satisfies both equations.

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3

−

3

t

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d

y

d

t

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t

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L

=

∫

√

3

0

√

(

3

−

3

t

2

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2

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(

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+

9

t

4

+

36

t

2

d

t

=

∫

√

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0

√

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+

18

t

2

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t

4

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t

=

∫

√

3

0

√

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3

+

3

t

2

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2

d

t

=

∫

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3

0

(

3

+

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t

2

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d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=6The arclength of a parametric curve can be found using the formula:  

L

=

∫

t

f

t

i

√

(

d

x

d

t

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2

+

(

d

y

d

t

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2

d

t

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x

and  

y

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It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

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d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

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L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

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t

2

+

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t

4

+

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d

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=

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√

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√

(

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2

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√

3

0

(

3

+

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2

)

d

t

=

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t

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3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=

6

√

3

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.    

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.

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