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2 years ago

Find the quotient.

1 answer:

3 0

Answer :

$B. 16 {a}^{2}c$

step-by-step explanation :

$48 {a}^{3}b{c}^{2} \div 3abc$

This can be rewritten as:

$\frac{ 48 {a}^{3}b {c}^{2} }{3abc}$

Now,

$\frac{48}{3}=16$

The law of indices states that:

$\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

It implies that, when dividing two expressions with the same bases, repeat one of the bases and subtract the exponents.

Therefore,

$\frac{ {a}^{3} }{a}= {a}^{3 - 1} = {a}^{2}$

$\frac{b}{b} = {b}^{1 - 1} = {b}^{0} = 1$

Note: Any non-zero number exponent zero is 1

Also

$\frac{ {c}^{2} }{c} = {c}^{2 - 1} = {c}^{1} = c$

Hence:

$\frac{ 48 {a}^{3}b {c}^{2} }{3abc} = 16 {a}^{2}c$

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Determine the m m m m m

iVinArrow [24]

Answer:

A,D,E,G=83 B,C,F=97

Step-by-step explanation:

5 0

2 years ago

PLEASE HELP

cricket20 [7]

You can use the cosine rule for this - you might have to change some of the letters around for it to make sense though. Hope this helps!

4 0

2 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

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:

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

7 0

2 years ago

Tanzania [10]

Answer:

c = 160 - 3(45)

Step-by-step explanation:

if c = the number of characters left;

160 = maximum number of characters that can be used in a text

3*45 = number of characters in sentences already used. (There are 3 sentences of 45 characters.)

(# of char left) = (max # char) - (# char already used) so:

c = 160 - 3*(45)

c = 160 - 135

c = 25

8 0

2 years ago

What fraction of a gallon is 3 pints?

Nitella [24]

There are 8 pints in 1 gallon, so 3 pints is 3/8 of a gallon

4 0

2 years ago

Read 2 more answers