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KIM [24]
4 years ago
9

Complete the pattern 5,10,30,60,180,__

Mathematics
2 answers:
nalin [4]4 years ago
7 0

Hi!

If the pattern starts off with 5 going to 10, then that would be multiplying by 2. Then, from 10 to 30 would be multiplying by 3. From 30 to 60 would be 2, and 60 to 180 would be 3. So, the pattern is multiply by 2, then multiply by 3, then by 2, then by 3, and so on. So, if we multiply 180 by 2, we get 360, the answer.

Hope this helps!

jok3333 [9.3K]4 years ago
3 0

270 because you can realize it starts going up by 30 so you had to multiply 3 × 30 to get 90 then add 90 to 180 and get 270

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Answer:

hello the table related to this question is missing attached below is the missing table

answer : An exponential function is much better fit because there is no fixed rate in the table

Step-by-step explanation:

An exponential function is much better fit because there is no fixed rate in the table attached but the values in the table keeps increasing exponentially, hence an Exponential function will be a better fit, But a linear function would be preferred only if there is a fixed rate

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Assuming the population variances are unknown and equal, you wish to test whether the Group 1 population mean is equal to the Gr
Fantom [35]

Answer:

n_1+n_2-2

Step-by-step explanation:

If we had 2 independent samples with population variances unknown and equal and we want to compare the population mean between the both groups, the test statistic will be:

t=\frac{x_1-x_2}{\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2}) } }

Where s_p^2=\frac{(n_1-1)*s_1^2+(n_2-2)*s_2^2}{n_1+n_2-2}

So, the t-student has n_1+n_2-2 degrees of freedom

Where x_1 and x_2 are the means of the groups, n_1 and n_2 are the size of the samples and s_1^2 and s_2^2 are the sample variances.

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3 years ago
If kn equals 29 what is cn
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4 years ago
Solve for b. C=b225 b=±5C√C b=±C√5 b=±5C b=±5C‾‾√
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Now if you ever need help a website you can use is cymath, i use it for my school and its how i found out your answer here, it has never been wrong for me, i'm just a student myself. Good Luck!
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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
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