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lesya692 [45]
3 years ago
11

Which ordered pair is in the solution set of the system of linear inequalities graphed below?

Mathematics
1 answer:
motikmotik3 years ago
5 0

Answer:

Graphed?

Step-by-step explanation:

You might be interested in
Determine the gradient of the straight line 2x-3y+9=0. Find the equation of the straight line through the origin which is perpen
frutty [35]

Answer: 3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

x

1

)3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

Explanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular lineExplanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular line

7 0
3 years ago
Read 2 more answers
On a piece of paper, graph this system of inequalities. Then determine which
Daniel [21]
The answer is C, Region A.
3 0
2 years ago
I used the pythagorean theorem.<br><br> Need a second opinion
uysha [10]

Answer:

A.

Step-by-step explanation:

EF^2=18^2-12^2

EF^2=324-144

EF^2=180

EF=root 180

7 0
3 years ago
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP
erik [133]

Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

-x^{2} +2x+10-x-2=0

-x^{2} +x+8=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

5 0
3 years ago
Which expression is equivalent to Cube root of 216 x cubed y Superscript 6 Baseline z Superscript 12?
yaroslaw [1]

Answer:

6xy²z^4

Step-by-step explanation:

\sqrt[3]{216x^3y^6z^12} = ∛(6³)(x³)(y³)(y³)(z³)(z³)(z³)(z³) = 6x·y·y·z·z·z·z = 6xy²z^4

3 0
3 years ago
Read 2 more answers
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