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3 years ago

When adding numbers can you move them around? A. yes B. no

Mathematics

1 answer:

oee [108]3 years ago

7 0

Answer:

yes

Step-by-step explanation:

The law lets us move all of the addends around in any addition problem.

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-k+0.03+1.01k=-2.45-1.81k?

Mrrafil [7]

Answer:

-1.362637

or if you round your answer -1.36

Step-by-step explanation:

solve for k by simplifying both sides of the equation and isolating the variable.

4 0

3 years ago

What is the value of x? Enter your answer in the box. x =

Triss [41]

Answer:

x = 4.1

Step-by-step explanation:

6x-3 + 3x+6 + 5x = 60

14x + 3 = 60

14x = 60 - 3

14x = 57

x = 57/14

x = 4.1

4 0

3 years ago

3/4 of a number is 24. Find the number.

Mice21 [21]

Answer:

Step-by-step explanation:

Divide 24 by 3/4 to find the number.

24 ÷ 3/4

= 24 × 4/3

= 32

8 0

2 years ago

Read 2 more answers

What is the value of 3n + 1 when = 4 ?

Jet001 [13]

Answer:

Step-by-step explanation:

Put n as 4

$3(4)+1=12+1=13$

Hope this helps :)

7 0

3 years ago

Read 2 more answers

The math question is on the image find the nth term of the sequence

Rus_ich [418]

Notice how Pattern 2 is Pattern 1 with 4 balls added in the bottom row.

Pattern 3 is Pattern 2 with 5 more balls.

Pattern 4 is Pattern 3 with 6 more balls.

Generalizing the trend, we expect Pattern $n$ to be identical to Pattern $n-1$ with $n+2$ more balls.

If $b_n$ is the number of balls in the $n$ -th pattern, then we have the recursive relation

$\begin{cases} b_1 = 6 \\ b_n = b_{n-1} + n + 2 & \text{for } n>1 \end{cases}$

We can solve this recurrence by substitution. Using the definition of $b_n$ , we have

$b_{n-1} = b_{n-2} + (n-1) + 2 \\\\ \implies b_n = (b_{n-2} + (n-1) + 2) + n + 2 \\\\ \implies b_n = b_{n-2} + 2\times2 + \bigg(n + (n-1)\bigg)$

$b_{n-2} = b_{n-3} + (n-2) + 2 \\\\ \implies b_n = (b_{n-3} + (n-2) + 2) + 2\times 2 + \bigg(n + (n-1)\bigg) \\\\ \implies b_n = b_{n-3} + 3\times2 + \bigg(n + (n-1) + (n-2)\bigg)$

and so on, down to

$b_n = b_1 + (n-1)\times2 + \bigg(n + (n-1) + (n-2) + \cdots + 2\bigg)$

Recall that

$\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

Then we find

$\displaystyle b_n = 6 + 2(n-1) + \sum_{i=2}^n i$

$\displaystyle b_n = 2n + 4 + \left(\sum_{i=1}^n i - 1\right)$

$\displaystyle b_n = 2n + 3 + \frac{n(n+1)}2$

$\displaystyle \boxed{b_n = \frac{n^2+5n+6}2}$

8 0

2 years ago