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IgorC [24]
3 years ago
8

5) What point would be the reflection, across the x-axis, for the point (-8,-7)?

Mathematics
1 answer:
melisa1 [442]3 years ago
8 0

Answer: (8,-7)

Step-by-step explanation:  Since you are reflecting over the x-axis, the x coordinate stays the same, but the y coordinate does change. Instead of it being negative, the eight would become positive.

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(2*)2-3×2*+2=0<br>4m-15°(×)m+75°​
Valentin [98]

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

7 0
2 years ago
What is the length of arc S?<br> (The angle in the figure is a central angle in radians)
IgorC [24]

Answer:

π/90 or 0.0035 units

Step-by-step explanation:

equation: <em>length of the arc = ∠of the angle/360 * circumference</em>

Substitute: <em>S = 0.4/360 * 10π</em> --> C = 2πr

Simplify: S = 1/900 * 10π

Simplify: S = π/90 ≈ 0.003489 units

7 0
3 years ago
(1 3/4)²<br><br> help please
Pie

Answer:

\frac{49}{16}  \: or \: 3 \frac{1}{16}

5 0
2 years ago
Describe in words the region of R3 reperesented by the equations or inequalities x2+y2=4,z=-1
Roman55 [17]

Answer:

It represents a circle of radius 2, centered in the origin of the plane xy and transversal to the z-axis in z=-1.

Step-by-step explanation:

The first thing to notice is that

x^2+y^2=2^2

<u>represents a circle of radius 2</u>, with its center in the origin of a plane xy, of cartesians coordinates.

Starting from here, we have to put the coordinate z, to complete the space R³. <em>Then, the cirlce will "live" into the plane xy, where z=-1</em>.

Finally, we have that this region is a circle of radius 2, centered in the origin of the plane xy, and transversal to z=-1.

8 0
2 years ago
What's the constant numeric value of 8b + 3.2<br><br><br><br> Pls answer quick
Svetach [21]

Answer:

the answer shouild be 8 if not them let me know and i can figure it out

Step-by-step explanation:

5 0
3 years ago
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