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3 years ago

Which of the following expressions is not equivalent to (-2)(8 +6+-3)?

Mathematics

1 answer:

kari74 [83]3 years ago

7 0

Answer:

$b.)31\neq -22$

Step-by-step explanation:

Simplify each equation using PEMDAS:

Parentheses (x)

Exponents x²

Multiplication and Division, left to right × ÷

Addition and Subtraction, left to right - +

It is also important to follow PEMDAS when working within grouping symbols like parentheses.

Also, remember that two negatives make a positive, and when a negative and positive are multiplied, the result is always negative.

$(-2)(8+6+-3)\\\\-2*(8+6-3)\\\\-2*(14-3)\\\\-2*11\\\\-22$

Find which answer is not equal to -22.

a.) $(-2)(8+6)+(-2)(-3)\\\\-2*(8+6)-2*-3\\\\-2*14-2*-3\\\\-28-2*-3\\\\-28+6\\\\-22$

---------------------------------------------------

b.) $(-2)*(8+6)+(-3)\\\\-2*(8+6)-3\\\\-2*14-3\\\\-28-3\\\\-31$

---------------------------------------------------

c.) $(-2)(8)+(-2)(6)+(-2)(-3)\\\\-2*8-2*6-2*-3\\\\-16-2*6-2*-3\\\\-16-12-2*-3\\\\-16-12+6\\\\-28+6\\\\-22$

---------------------------------------------------

d.) $(-2)(8)+(-2)(6+-3)\\\\-2*8-2*(6-3)\\\\-2*8-2*3\\\\-16-2*3\\\\-16-6\\\\-22$

---------------------------------------------------

Option b is not equal to -22.

:Done

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Lim (n/3n-1)^(n-1) n → ∞

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Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$