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$32,520 divided by 30 people

Ronch [10]

Answer: $1,084 per person

Step-by-step explanation:

divide 32520 by 30

8 0

3 years ago

Which graph shows the solution to the inequality -3x-7 <20

IgorLugansk [536]

Well I would give you the answer but I can not because there are no graphs for me to pick from

4 0

3 years ago

Tracy wants to earn at least 100 dollars from her two jobs next week. At most she can work 12 hours. Her first job pays 8 an hou

exis [7]

Answer:

Step-by-step explanation:

You didn't list the options from which we are to choose as your system of inequalities, but that doesn't matter...we'll come up with them on our own and then you can match them to your options. The first inequality is going to be about the number of hours worked. The second inequality is going to be about the money earned. Hours worked and money earned have to be in 2 different inequalities because they are not the same. If x is one job and y is the other, and the combination of these jobs cannot be more than 12 hours total, then the inequality for this is:

x + y ≤ 12

That represents the hours worked. As far as the money goes, she makes $8 per hour, x, at the first job, and $9 per hour, y, at the second job. She wants the combination of these wages to be at least $100. The inequality that represents the money earned is:

8x + 9y ≥ 100

That is the system that represents your situation.

7 0

3 years ago

Use the long division method to find the result when x^3+9x² +21x +9 is divided<br> by x+3

Serhud [2]

Answer:

x^3 + 9 x^2 + 21 x + 9 = (x^2 + 6 x + 3)×(x + 3) + 0

Step-by-step explanation:

Set up the polynomial long division problem with a division bracket, putting the numerator inside and the denominator on the left:

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

To eliminate the leading term of the numerator, x^3, multiply x + 3 by x^2 to get x^3 + 3 x^2. Write x^2 on top of the division bracket and subtract x^3 + 3 x^2 from x^3 + 9 x^2 + 21 x + 9 to get 6 x^2 + 21 x + 9:

| | | x^2 | | | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

To eliminate the leading term of the remainder of the previous step, 6 x^2, multiply x + 3 by 6 x to get 6 x^2 + 18 x. Write 6 x on top of the division bracket and subtract 6 x^2 + 18 x from 6 x^2 + 21 x + 9 to get 3 x + 9:

| | | x^2 | + | 6 x | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

To eliminate the leading term of the remainder of the previous step, 3 x, multiply x + 3 by 3 to get 3 x + 9. Write 3 on top of the division bracket and subtract 3 x + 9 from 3 x + 9 to get 0:

| | | x^2 | + | 6 x | + | 3

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

| | | | | -(3 x | + | 9)

| | | | | | | 0

The quotient of (x^3 + 9 x^2 + 21 x + 9)/(x + 3) is the sum of the terms on top of the division bracket. Since the final subtraction step resulted in zero, x + 3 exactly divides x^3 + 9 x^2 + 21 x + 9 and there is no remainder.

| | | x^2 | + | 6 x | + | 3 | (quotient)

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9 |

| -(x^3 | + | 3 x^2) | | | | |

| | | 6 x^2 | + | 21 x | + | 9 |

| | | -(6 x^2 | + | 18 x) | | |

| | | | | 3 x | + | 9 |

| | | | | -(3 x | + | 9) |

| | | | | | | 0 | (remainder) invisible comma

(x^3 + 9 x^2 + 21 x + 9)/(x + 3) = (x^2 + 6 x + 3) + 0

Write the result in quotient and remainder form:

Answer: Set up the polynomial long division problem with a division bracket, putting the numerator inside and the denominator on the left:

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

To eliminate the leading term of the numerator, x^3, multiply x + 3 by x^2 to get x^3 + 3 x^2. Write x^2 on top of the division bracket and subtract x^3 + 3 x^2 from x^3 + 9 x^2 + 21 x + 9 to get 6 x^2 + 21 x + 9:

| | | x^2 | | | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

To eliminate the leading term of the remainder of the previous step, 6 x^2, multiply x + 3 by 6 x to get 6 x^2 + 18 x. Write 6 x on top of the division bracket and subtract 6 x^2 + 18 x from 6 x^2 + 21 x + 9 to get 3 x + 9:

| | | x^2 | + | 6 x | |

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

To eliminate the leading term of the remainder of the previous step, 3 x, multiply x + 3 by 3 to get 3 x + 9. Write 3 on top of the division bracket and subtract 3 x + 9 from 3 x + 9 to get 0:

| | | x^2 | + | 6 x | + | 3

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9

| -(x^3 | + | 3 x^2) | | | |

| | | 6 x^2 | + | 21 x | + | 9

| | | -(6 x^2 | + | 18 x) | |

| | | | | 3 x | + | 9

| | | | | -(3 x | + | 9)

| | | | | | | 0

The quotient of (x^3 + 9 x^2 + 21 x + 9)/(x + 3) is the sum of the terms on top of the division bracket. Since the final subtraction step resulted in zero, x + 3 exactly divides x^3 + 9 x^2 + 21 x + 9 and there is no remainder.

| | | x^2 | + | 6 x | + | 3 | (quotient)

x + 3 | x^3 | + | 9 x^2 | + | 21 x | + | 9 |

| -(x^3 | + | 3 x^2) | | | | |

| | | 6 x^2 | + | 21 x | + | 9 |

| | | -(6 x^2 | + | 18 x) | | |

| | | | | 3 x | + | 9 |

| | | | | -(3 x | + | 9) |

| | | | | | | 0 | (remainder) invisible comma

(x^3 + 9 x^2 + 21 x + 9)/(x + 3) = (x^2 + 6 x + 3) + 0

Write the result in quotient and remainder form:

Answer: x^3 + 9 x^2 + 21 x + 9 = (x^2 + 6 x + 3)×(x + 3) + 0

5 0

2 years ago

How to prove all pairs of corresponding sides and angles are congruent

Lera25 [3.4K]

The lines on the triangles

5 0

3 years ago