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3 years ago

Please help; I don't understand how to solve these inequality problems.

Mathematics

2 answers:

jarptica [38.1K]3 years ago

6 0

Answer:

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

Darina [25.2K]3 years ago

3 0

Answer:

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.

-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

6 - 2x ≤ 12 or 7 + 2x < -11

Disjunction

-2x ≤ 6 or 2x < -18

x ≥ -3 or x < -9

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3 years ago

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WINSTONCH [101]

Answer:

The answer is hyperbola; (x')² - (y')² - 16 = 0 ⇒ answer (a)

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

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the origin by angle Ф

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∴ We rotated the x-axis and the y-axis by angle π/4

* That means the point (x' , y') it was point (x , y)

- Where x' = xcosФ - ysinФ and y' = xsinФ + ycosФ

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∴ $(\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=$

( $\frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0$

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∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

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olya-2409 [2.1K]

<h3>Answer: 5 cakes</h3>

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Explanation:

Let's start off converting the mixed number 12 & 1/4 to an improper fraction.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\12 \frac{1}{4} = \frac{12*4+1}{4}\\\\12 \frac{1}{4} = \frac{49}{4}\\\\$

Do the same for the other mixed number 2 & 1/3.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\2 \frac{1}{3} = \frac{2*3+1}{3}\\\\2 \frac{1}{3} = \frac{7}{3}\\\\$

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From here, we divide the two fractions. I converted them to improper fractions to make the division process easier.

$\frac{49}{4} \div \frac{7}{3} = \frac{49}{4} \times \frac{3}{7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{49\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 7\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 3}{4}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{21}{4}\\\\$

The last step is to convert that result to a mixed number.

$\frac{21}{4} = \frac{4*5+1}{4}\\\\\frac{21}{4} = \frac{4*5}{4}+\frac{1}{4}\\\\\frac{21}{4} = 4+\frac{1}{4}\\\\\frac{21}{4} = 5 \frac{1}{4}\\\\$

Note that 21/4 = 5.25 and 1/4 = 0.25 to help check the answer.

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Therefore, she can make 5 cakes. The fractional portion 1/4 is ignored since we're only considering whole cakes rather than partial ones.

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