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Archy [21]
2 years ago
14

Value of x in the given figure​

Mathematics
2 answers:
Natasha2012 [34]2 years ago
8 0

Answer:

x = 3

Step-by-step explanation:

To find the value of x we need to

(6x - 7) = (4x - 1) <em>(VERTICALLY OPPOSITE ANGLES ARE EQUAL)</em>

Transpose

6x -4x = -1 +7

2x = 6

x = 6 ÷ 2

x = 3

ololo11 [35]2 years ago
6 0

Answer:

\sf\longmapsto \: x = 3

Step-by-step explanation:

\sf\longmapsto \: 6x  - 7 = 4x - 1

\sf\longmapsto \: 6x - 4x =  - 1 + 7

\sf\longmapsto \: 2x = 6

\sf\longmapsto \: x = 6 \div 2

\sf\longmapsto \: x = 3

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In quadrilateral DBCA, name the angles which are consecutive with angle D.
lisov135 [29]

Answer: The consecutive angles with angle D are B and A.

Explanation:

It is given that DBCA is a quadrilateral. Since it is a quadrilateral it means the have 4 vertices and 4 angles D, B,C,A.

The angles are formed in the order of the figure name. If the quadrilateral name is DBCA, So the order of the angles are D,B,C,A. It means the angle immediate after D is B, the angle immediate after B is C, the angle immediate after C is A and the angle immediate after A is D.

If the quadrilateral name is DBCA it means the sides are DB, BC, CA and AD as shown in the figure.

Consecutive angles of D means the angle immediate before and after the angle D.

In the figure there are some types of quadrilateral and from the figure we can easily noticed that the consecutive angles with angle D are B and A.

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3 years ago
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A bank paid 2.3% interest on a money market account. What is 2.3% written as a decimal
DochEvi [55]
1\%\ \ \rightarrow\ \  \frac{1}{100}=0.01 \\\\2.3\%\ \ \rightarrow\ \ x\\\\x= \frac{2.3\%\cdot0.01}{1\%} \ \ \ \Rightarrow\ \ \ x=0.023
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3 years ago
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May I please receive help?
Jlenok [28]

Answer:

The answer is in the included image

7 0
3 years ago
Help me on this please
zalisa [80]

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each <u><em>IMAGE</em></u> you should perform the following steps:

  • List the vertex points of the preimage (the original figure) as ordered pairs.
  • Apply the transformation rule to every point of the preimage
  • List the image of each vertex after applying each transformation, also as ordered pairs.

<u>1. (x, y) → (x + 3, y - 2)</u>

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage      Vertices of the image

a) (-5,2)                                   (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5)                                  (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4)                                   (-1 + 3, 4 - 2) = (2, 2)

<u>2. (x,y) → (x - 3, y + 5)</u>

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage      Vertices of the image

a) (0, -3)                                   (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3)                                   (3 - 3, -3  + 5) = (0, 2)

c) (3, -1)                                    (3 - 3, -1 + 5) = (0, 4)

d) (5, -1)                                    (5 - 3, -1 + 5) = (2, 4)

<u>3. (x, y) → (x + 4, y)</u>

The rule represents a translation 4 units to the right.

Vertices of the preimage   Vertices of the image

a) (-5, -2)                               (-5 + 4, -2) = (-1, -2)

b) (-3, -5)                               (-3 + 4, -2) = (1, -2)

c) (-1, -2)                                (-1 + 4, -2) = (3, -2)

<u>4. (x, y) → (x + 6, y + 1)</u>

Vertices of the preimage      Vertices of the image

a) (-5, -2)                                  (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3)                                  (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3)                                   (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5)                                  (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2)                                  (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4)                                   (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4)                                  (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5)                                  (-2 + 3, -5 + 1) = (1, -4)

<u>5. (x, y) → (x, y - 4)</u>

This is a translation four units down

Vertices of the preimage      Vertices of the image

a) (0, 2)                                    (0, 2 - 4) = (0, -2)

b) (0,1)                                      (0, 1 - 4) = (0, -3)

c) (2, 2)                                     (2, 2 - 4) = (2, -2)

d) (2,0)                                     (2, 0 - 4) = (2, -4)

<u>6. (x, y) → (x - 1, y + 4)</u>

This is a translation one unit to the left and four units up.

Vertices of the pre-image     Vertices of the image

a) (-4, -1)                                   (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5)                                  (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4)                                  (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5)                                 (-2 - 1, -5 + 4) = (-3, -1)

8 0
3 years ago
Point D is a fixed point on line PQ such that PD is =DQ, the cordinate of P and Q are (2,3) and (4,7)
Savatey [412]

Answer:

2

Step-by-step explanation:

hope this helps

5 0
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