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3 years ago

Find the product. (2n + 2)(2n – 2)

2 answers:

7 0

Use the FOIL method, or (First, Outside, Inside, Last)

2n x 2n = 4n^2
2n x (-2) = -4n
2n x (2) = 4n
2 x (-2) = -4

4n² - 4n + 4n - 4 (simplify)

4n² (-4n + 4n) - 4

-4n + 4n = 0

4n² - 4 is your answer

hope this helps

vekshin13 years ago

3 0

<span>(2n + 2)(2n – 2)
= (2n)^2 -2^2
= 4n^2 - 4

hope it helps</span>

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A= [2, 3; 2, 1], B=[3; 5]. Find AB & BA if possible

oksano4ka [1.4K]

Answer:

The value of AB is $\left[\begin{array}{ccc}21\\11\end{array}\right]$ and it's not possible to multiply BA.

Step-by-step explanation:

Consider the provided matrices.

$A=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right]$ , $B=\left[\begin{array}{ccc}3\\5\end{array}\right]$

Two matrices can be multiplied if and only if first matrix has an order m × n and second matrix has an order n × v.

Multiply AB

Matrix A has order 2 × 2 and matrix B has order 2 × 1. So according to rule we can multiply both the matrix as shown:

$AB=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] \left[\begin{array}{ccc}3\\5\end{array}\right]$

$AB=\left[\begin{array}{ccc}2\times 3+3\times 5\\2\times 3+1\times 5\end{array}\right]$

$AB=\left[\begin{array}{ccc}6+15\\6+5\end{array}\right]$

$AB=\left[\begin{array}{ccc}21\\11\end{array}\right]$

Hence, the value of AB is $\left[\begin{array}{ccc}21\\11\end{array}\right]$

Now calculate the value of BA as shown:

Multiply BA

Matrix B has order 2 × 1 and matrix A has order 2 × 2. So according to rule we cannot multiply both the matrix.

We can multiply two matrix if first matrix has an order m × n and second matrix has an order n × v.

That means number of column of first matrix should be equal to the number of rows of second matrix.

Hence, it's not possible to multiply BA.

5 0

3 years ago

Round 399,411 to the nearest hundredth thousandth

melamori03 [73]

The nine rounds up so it would be 400,000

6 0

3 years ago

20 POINTS PLEAE HELP!!!!!!!!!!!!!!!

Alina [70]

His first payment is $100, thus a₁ = 100.

the next "term", month will be 1.1 times more than the one before, namely r = 1.1, the common ratio.

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=100\\
r=1.1\\
n=20
\end{cases}$

$\bf \sum\limits_{i=1}^{20}~100(1.1)^{i-1}\qquad \qquad\qquad \qquad S_{20}=100\left( \cfrac{1-1.1^{20}}{1-1.1} \right)
\\\\\\
S_{20}=100\left( \cfrac{1-\stackrel{\approx}{6.727499949}}{-0.1} \right)\implies S_{20}\approx 100(57.27499949)
\\\\\\
S_{20}\approx 5727.4999493256$

is the serie divergent or convergent?

well, to make it short, when the common ratio is 0 < | r | < 1, namely a fraction between 0 and 1, only then the serie is convergent, namely it reaches a fixed value, now in this case, 1.1 is a value larger than anything between 0 and 1, so no dice.

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3 years ago

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If you begin a problem with a one digit integer , multiply the number by 4 add 7 divide by 3 subtract 5 and answer of 0 what is

12345 [234]

Answer: imma figure this out

Step-by-step explanation:

6 0

3 years ago

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If you get it right ill give you brainlist

Shkiper50 [21]