;i don't understand how to solve these inequality problems.

for each problem:
solve the combined inequality.
explain step by step how you solved it.
state whether the combined inequality is a conjunction or disjunction and explain.
determine which letter is in the solution.

1. -4≤x+2< 1

2. 5x-4> 46" or " 4x≤3x+6

3. 10≤2x-4≤20

4. 6-2x≤12" or " 7+2x< -11

5. 5≤2x+1≤11

6. 2/3 x-20≤16" or " x/3+10≥32

7. 10< 1/4 (x+1)≤11

1. -6 ≤ x < -1, conjunction

2. x > 10   or   x ≤ 6, disjunction

3. 7 ≤ x ≤ 12, conjunction

4. x ≥ -3   or   x < -9, disjunction

Step-by-step explanation:

These inequalities are called "compound inequalities." Each compound inequality is made up of two simple inequalities.

A compound inequality of the type 5 < x < 8 means x > 5 and x < 8. Since the word between the simple inequalities is "and", it is a conjunction.

A compound inequality of the type "x < 3 or x > 12" uses the word "or" between the simple inequalities. It is called a disjunction.

1.

-4 ≤ x + 2 < 1

Conjunction

For this type of inequality, do what you need to do to get x alone in the middle section. Do the same to all three "sides" of the inequality.

The middle section has x + 2. We want x alone,s o w must subtract 2. We subtract 2 from all three sides.

-4 - 2 ≤ x + 2 - 2 < 1 - 2

-6 ≤ x < -1

2. 5x - 4 > 46 or 4x ≤ 3x + 6

Disjunction

In this type of compound inequality, solve each inequality by itself, and always keep the word "or" between the inequalities.

5x - 4 > 46 or 4x ≤ 3x + 6

5x - 4 + 4 > 46 + 4   or   4x - 3x ≤ 3x - 3x + 6

5x > 50   or   x ≤ 6

x > 10   or   x ≤ 6

3. Similar to problem 1.

Conjunction

10 ≤ 2x - 4 ≤ 20

Add 4 to all sections.

14 ≤ 2x ≤ 24

Divide all sections by 2.

7 ≤ x ≤ 12

4.

6 - 2x ≤ 12 or 7 + 2x < -11

Disjunction

-2x ≤ 6   or   2x < -18

x ≥ -3   or   x < -9

1. -6≤x<-1

conjunction

2.x>10   or     x≤6

Disjunction

3.7≤x≤12

Conjunction    

4.x ≥-6                    or                           x < -6

Disjunction

5.2≤x≤5

Conjunction

6. x ≤ 54                                        or               x≥66

Disjunction

7.  39< x≤43

Conjunction

Step-by-step explanation:

conjunction is "and"  like a sandwich between 2 points

disjunction is " or"  like boat oars with a gap for the boat in the middle

1. -4≤x+2<1

Subtract 2 from all sides

-4-2≤x+2-2<1 -2

-6≤x<-1

conjunction

2. 5x-4>46"                                 or          " 4x≤3x+6

Add 4 to each side                                  Subtract 3x from each side

5x-4+4 > 46+4                                           4x-3x≤3x-3x+6

5x >50                                                           x≤6

Divide by 5                                                  

5x/5 >50/5

x>10   or     x≤6

Disjunction

3. 10≤2x-4≤20

Add 4 to all sides

10≤2x-4≤20

10+4 ≤2x-4+4≤20+4

14≤2x≤24

Divide by 2 on all sides

14/2≤2x/2≤24/2

7≤x≤12

Conjunction    

4. 6-2x≤12" or                             " 7+2x<-11

Subtract 6 from each side       Subtract 7 from each side

6-6-2x≤12                                or " 7-7+2x<-11-7

-2x≤12                                        2x<-18

Divide by -2                                Divide by 2

Flips the inequality

-2x/-2≥12/-2                             2x/2 <-18/2

x ≥-6                    or                           x < -6

Disjunction

5. 5≤2x+1≤11

Subtract 1 from all sides

5-1≤2x+1-1≤11-1

4≤2x≤10

Divide all sides by 2

4/2≤2x/2≤10/2

2≤x≤5

Conjunction

6. 2/3 x-20≤16"                            or " x/3+10≥32

Add 20 from each side                     Subtract 10 from each side

2/3 x-20+20 ≤16+20                     x/3 +10-10 ≥32 -10

2/3 x ≤36                                                x/3≥22

Multiply by 3/2 on each side            Multiply by 3 on each side

3/2 * 2/3x ≤ 36*3/2                      x/3*3 ≥22 *3

x ≤ 54                                        or               x≥66

Disjunction

7. 10<1/4 (x+1)≤11

Multiply all sides by 4

4*10<1/4*4 (x+1)≤11*4

40< (x+1)≤44

Subtract 1 from all sides

40-1< (x+1-1)≤44-1  

39< x≤43

Conjunction

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204