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3 years ago

What is the common difference for this arithmetic sequence?

Mathematics

3 answers:

olya-2409 [2.1K]3 years ago

7 0

Subtract a number from the next number.

39 - 45 = -6

Answer: B. -6

Musya8 [376]3 years ago

5 0

B. -6
The numbers in the equation are decreasing meaning that the common difference must be a negative number.

Nettjordan2 years ago

0 0

2749 is my answer now now

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What is 3 minus 3 times 6 plus 2?

Zina [86]

What is 3 minus 3 times 6 plus 2?

Write it out:-

3 - 5 × 6 + 2 = ?
3 - 30 + 2 = ?
-27 + 2 =
-25

<span>3 minus 3 times 6 plus 2 is -25. </span>

7 0

3 years ago

Read 2 more answers

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

6 0

2 years ago

How do I solve these for example 3

Natasha2012 [34]

As the three longitudinal lines are parallel:

angle 4 = 5 = 6 = 82 degree (corresponding angles) = 9 = 8 = 7 = 82 degrees (vertical angles)

and

angle 11 = 180-82= 98 degrees (linear pair)

so:

angle 10 = 11 = 12 = 98 degrees (corresponding angles) = 3 = 2 = 1 = 98 degrees (vertical angles)
.

hope you got it

8 0

3 years ago

Solve for equation x. <br> 3In(x)+2In(4)=In(128)

Serggg [28]

$Domain:\ x > 0\\\\3\ln(x)+2\ln(4)=\ln(128)\qquad\text{use}\ \lag_ab^n=n\log_ab\\\\\ln(x^3)+\ln(4^2)=\ln(128)\qquad\text{use}\ \log_ab+\ln_ac=\ln_a(bc)\\\\\ln(16x^3)=\ln(128)\iff16x^3=128\qquad\text{divide both sides by 16}\\\\x^3=8\to x=\sqrt[3]8\\\\\boxed{x=2}\in D$

3 0

2 years ago

The base of a triangular pyramid has an area of 20 square feet. The height of the pyramid is 36 feet. The volume of the pyramid

fenix001 [56]

Answer:

so what's the question? i'm a bit confused..

Step-by-step explanation:

7 0

2 years ago