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rodikova [14]
3 years ago
10

Which are correct representations of the inequality 6x ≥ 3 + 4(2x – 1)? Check all that apply.

Mathematics
2 answers:
Verdich [7]3 years ago
6 0

Answer:

6x ≥ 3 + 4(2x - 1)

⇔ 6x ≥ 3 + 8x - 4        => remove the parentheses

⇔ 6x - 8x ≥ 3 - 4

⇔ -2x ≥ -1

⇔ 2x ≤ 1 (or 1 ≥ 2x)

⇔ x ≤ 1/2

⇔ x ≤ 0.5

The answer for the question should be:

1 ≥ 2x

6x ≥ 3 + 8x – 4

and the first graph.

Anika [276]3 years ago
5 0

Answer:

1 ≥ 2x

6x  ≥ 3 + 8x - 4

and the first line graph.

Step-by-step explanation:

6x ≥ 3 + 4(2x – 1)

6x  ≥ 3 + 8x - 4

6x -  8x ≥  -1

-2x   ≥ -1

x  ≤ 1/2

or 1 ≥ 2x.

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gtnhenbr [62]
It is C) 1/25
Hope this helps
7 0
3 years ago
Find the area of the parallelogram that has a base of 4m and a height of 5.5m.
Sveta_85 [38]

Answer:

22m²

Step-by-step explanation:

Area of parallelogram is simply the base length x height.

In this case, base length = 4m and height = 5.5m

hence,

Area = 4 x 5.5 = 22m²

3 0
3 years ago
Read 2 more answers
Find the solution to the system 4x - y = -12 and -x - y = 3
nikdorinn [45]

Answer:

x = -3 , y = 0

Step-by-step explanation:

Solve the following system:

{4 x - y = -12 | (equation 1)

-x - y = 3 | (equation 2)

Add 1/4 × (equation 1) to equation 2:

{4 x - y = -12 | (equation 1)

0 x - (5 y)/4 = 0 | (equation 2)

Multiply equation 2 by 4/5:

{4 x - y = -12 | (equation 1)

0 x - y = 0 | (equation 2)

Multiply equation 2 by -1:

{4 x - y = -12 | (equation 1)

0 x+y = 0 | (equation 2)

Add equation 2 to equation 1:

{4 x+0 y = -12 | (equation 1)

0 x+y = 0 | (equation 2)

Divide equation 1 by 4:

{x+0 y = -3 | (equation 1)

0 x+y = 0 | (equation 2)

Collect results:

Answer: {x = -3 , y = 0

8 0
3 years ago
What’s the answer to this
madam [21]
B is the correct answer I think
4 0
3 years ago
Read 2 more answers
6x − y + z = −3<br> 4x − 3z = −13<br> 2y + 5z = 15<br> Find the solutions?
Alla [95]

Answer:

The solution of the equation are x = - 1 , y = 0 and z = 3

Step-by-step explanation:

Given three linear equation as :

6 x - y + z = - 3        ......A

4 x - 3 z = - 13           ......B

2 y + 5 z = 15            ......C

Solving eq A and C

I.e 2× (6 x - y + z ) = -3 ×2

or, 12 x - 2 y + 2 z = - 6

So, ( 12 x - 2 y + 2 z ) + ( 2 y + 5 z) = - 6 + 15

Or,  12 x + 7 z = 9         ......D

Solving eq B and D

I.e 3 × ( 4 x - 3 z ) = - 13 × 3

or, 12 x - 9 z = - 39

So, (   12 x + 7 z ) - ( 12 x - 9 z ) = 9 + 39

or,  16 z = 48

∴   z = \frac{48}{16}

i.e  z = 3

put the value of z in eq D

So, 12 x + 7×3 = 9  

Or, 12 x = 9 - 21

or,  12 x = - 12

∴   x = -  \frac{12}{12}

I.e  x = - 1

Now, Put The value of z in eq C

or, 2 y + 5 z = 15    

or,  2 y + 5 × 3 = 15

Or,  2 y = 15 - 15

or,  2 y = 0  

∴  y = 0

Hence The solution of the equation are x = - 1 , y = 0 and z = 3  Answer

8 0
3 years ago
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