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ElenaW [278]
3 years ago
10

Find the value of x a. 18 b. 7 c. 25 d. 11

Mathematics
1 answer:
Gwar [14]3 years ago
4 0
B is the answer if I’m correct
You might be interested in
Please help it's me! ​
Sever21 [200]

9514 1404 393

Answer:

  5

Step-by-step explanation:

The left side simplifies to ...

  2x +16

The right side simplifies to ...

  2x +21 -__

The x-terms are already equal. To make the constant terms equal, we want ...

  16 = 21 -__

The blank must be filled with the value 21-16 = 5 to make the statement true.

5 goes in the blank.

6 0
3 years ago
Cami and Enrique's ages add up to 46. Cami is two years older than three times Enrique's age. How old are Cami and Enrique?
stira [4]

answer:

Cami is 35 and Enrique is 11

35 plus 11 = 46

3 0
3 years ago
(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
Help this is the last question to a test
goblinko [34]
You’re a Horrible Person
Solutions are given in alphabetical order (k, x, y, z)

[(1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0), (2, 0, 1, 1), (2, 1, 0, 1), (2, 1, 1, 0), (3, 1, 1, 1), (8, 0, 0, 2), (8, 0, 2, 0), (8, 2, 0, 0), (9, 0, 1, 2), (9, 0, 2, 1), (9, 1, 0, 2), (9, 1, 2, 0), (9, 2, 0, 1), (9, 2, 1, 0), (10, 1, 1, 2), (10, 1, 2, 1), (10, 2, 1, 1), (16, 0, 2, 2), (16, 2, 0, 2), (16, 2, 2, 0), (17, 1, 2, 2), (17, 2, 1, 2), (17, 2, 2, 1), (24, 2, 2, 2), (27, 0, 0, 3), (27, 0, 3, 0), (27, 3, 0, 0), (28, 0, 1, 3), (28, 0, 3, 1), (28, 1, 0, 3), (28, 1, 3, 0), (28, 3, 0, 1), (28, 3, 1, 0), (29, 1, 1, 3), (29, 1, 3, 1), (29, 3, 1, 1), (35, 0, 2, 3), (35, 0, 3, 2), (35, 2, 0, 3), (35, 2, 3, 0), (35, 3, 0, 2), (35, 3, 2, 0), (36, 1, 2, 3), (36, 1, 3, 2), (36, 2, 1, 3), (36, 2, 3, 1), (36, 3, 1, 2), (36, 3, 2, 1), (43, 2, 2, 3), (43, 2, 3, 2), (43, 3, 2, 2), (54, 0, 3, 3), (54, 3, 0, 3), (54, 3, 3, 0), (55, 1, 3, 3), (55, 3, 1, 3), (55, 3, 3, 1), (62, 2, 3, 3), (62, 3, 2, 3), (62, 3, 3, 2), (64, 0, 0, 4), (64, 0, 4, 0), (64, 4, 0, 0), (65, 0, 1, 4), (65, 0, 4, 1), (65, 1, 0, 4), (65, 1, 4, 0), (65, 4, 0, 1), (65, 4, 1, 0), (66, 1, 1, 4), (66, 1, 4, 1), (66, 4, 1, 1), (72, 0, 2, 4), (72, 0, 4, 2), (72, 2, 0, 4), (72, 2, 4, 0), (72, 4, 0, 2), (72, 4, 2, 0), (73, 1, 2, 4), (73, 1, 4, 2), (73, 2, 1, 4), (73, 2, 4, 1), (73, 4, 1, 2), (73, 4, 2, 1), (80, 2, 2, 4), (80, 2, 4, 2), (80, 4, 2, 2), (81, 3, 3, 3), (91, 0, 3, 4), (91, 0, 4, 3), (91, 3, 0, 4), (91, 3, 4, 0), (91, 4, 0, 3), (91, 4, 3, 0), (92, 1, 3, 4), (92, 1, 4, 3), (92, 3, 1, 4), (92, 3, 4, 1), (92, 4, 1, 3), (92, 4, 3, 1), (99, 2, 3, 4), (99, 2, 4, 3), (99, 3, 2, 4), (99, 3, 4, 2), (99, 4, 2, 3), (99, 4, 3, 2)]
3 0
3 years ago
Given: AE ≅ CE ; DE ≅ BE Prove: ABCD is a parallelogram. Parallelogram A B C D is shown. Diagonals are drawn from point A to poi
nikklg [1K]

Answer:

The proof is below

Step-by-step explanation:

Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.

In ΔACD and ΔBEC

AD=BC              (∵Opposite sides of a parallelogram are equal)

∠DAC=∠BCE       (∵Alternate angles)

∠ADC=∠CBE        (∵Alternate angles)

By ASA rule, ΔACD≅ΔBEC

By CPCT(Corresponding Parts of Congruent triangles)

AE=EC and DE=EB

Hence, AE is conruent to CE and BE is congruent to DE

6 0
3 years ago
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