Can someone plz help me with this one problem plzzzzz!!! I will mark brainliest

a total of 408 tickets were sold for the school play. They were either adult tickets or student tickets. The number of students

Mama L [17]

Given Situation:
=> There are 408 total of ticket sold for a school play
=> 3 times of the tickets are sold to students compare to the adults
=> So the ratio of the given situation is 1 : 3
=> Now, 1 + 3 = 4
Since, we need 4 parts, let’s divide 408 by 4
=> 408 / 4 = 102 – the number of tickets for adults
=> 102 x 3 = 306 – the number of tickets for students
=> 102 : 306

4 0

3 years ago

10/3+7) i need help anyone got the answer ?

brilliants [131]

Answer:

Step-by-step explanation:

$\frac{10}{3}+7$ or $\frac{10}{3+7}$

If it is the former then first we will divide 10 by 3.

You get 3.333333

From there you will add 7 to that. You get 10.333333

If it is the latter then the answer is 1.

I know this because, you will first want to add 3 and 7. You get, 10.

Then you divide 10 by 10. Now you get 1.

Thus, the answer is 1.

Hope this helped! :)

4 0

3 years ago

A sphere has a diameter of 34 ft. What is its surface area?

Mars2501 [29]

Answer:

≈ 3629.84 ft^2

Step-by-step explanation:

Surface Area = 4πr^2

radius: 17 ft

= 4 · 3.14 · 17^2

= 4 · 3.14 · 289

≈ 3629.84 ft^2

4 0

3 years ago

Why is this number not in scientific notation? 36.5´ 106

love history [14]

Answer:

36.5 = 3.65x10^1

106= 1.06x10^2

6 0

4 years ago

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