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2 years ago

Kenia has purchased a water filter for her household. As per contract, Kenia

Mathematics

1 answer:

nadezda [96]2 years ago

3 0

Well I really don’t know but good luck w this I hope you pass school is so stressful

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Which algebraic properties was applied to the expression a + (b + c) = (a + b) + c?

andrew11 [14]

Answer:

the Associative Property

Step-by-step explanation:

Notice that the ORDER doesn't change -- a, b, c on both sides. But the GROUPING changes.

Example:

3 + (4 + 5) = (3 + 4) + 5

3 + 9 = 7 + 5

12 = 12

Subtraction does NOT behave this way!

8 - (7 - 1) = 8 - 6 = 2

(8 - 7) - 1 = 1 - 1 = 0

It matters who you associate with! :-)

5 0

2 years ago

What number has a place value of that is 100 times the place value of 6 in 2,386?

aivan3 [116]

Answer:

1,000

Step-by-step explanation:

It would be 3 places to the left, that is 1,000 so ya!

3 0

3 years ago

Read 2 more answers

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

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

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

What is ⅓ (1 third) times three??

RoseWind [281]

The answer to this problem is one because 1/3 3 times is 3/3 which is equal to one.

3 0

2 years ago

Read 2 more answers

(03.05 MC) Solve the rational equation x divided by 2 equals x squared divided by quantity x minus 2 end quantity, and check for

ycow [4]

Answer:

x = 0 and x = -2 are solutions of the given rational equation.

Step-by-step explanation:

We must solve the following rational equation:

$\frac{x}{2} = \frac{x^{2}}{x-2}$

Now we present the procedure:

1) $\frac{x}{2} = \frac{x^{2}}{x-2}$ Given

2) $x\cdot (x-2) = 2\cdot x^{2}$ Compatibility with multiplication/Existence of the multiplicative inverse/Definition of division/Modulative property.