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2 years ago

Melissa has more than 40 coins in her collection which inequality the nuber of coinsin melissas collection

Mathematics

1 answer:

nata0808 [166]2 years ago

4 0

Answer:

X > 40

Step-by-step explanation:

Given that:

Let number of coins in her collection is more Than 40

To represent the number of coins in Melissa's collection as an inequality :

Let total number of items in her collection = X

Coins in her collection > 40

Hence, coins in her collection can be represented as :

X > 40

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After a snowball fight, Chad made 4 mugs of hot chocolate to help everyone warm up. He had 1/2 of a cup of marshmallows and put

kirill115 [55]

Answer:

1/8th cup of marshmallows in each mug

Step-by-step explanation:

In order to split up the 1/2 cup of marhsmallows into 4 cups, you have to divide.

To divide, 1/2 by 4, you have to multiply 1/2 by the reciprocal of 4, which is 1/4.

1/2 x 1/4 = 1/8

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2 years ago

Read 2 more answers

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

Write the following numbers in order of size start with the smallest. 0.32 0.4 0.35 0.309

fenix001 [56]

Answer:

Hey!

0.309<0.32<0.35<0.4

hope this helps you

6 0

2 years ago

A mini-cone has a diameter of 3 inches and holds 10 in3 of ice cream. A mega-cone holds 49.13 in3 of ice cream. If the cones hav

marissa [1.9K]

The answer would be D

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2 years ago

Evaluate 12∣∣−x+3∣∣−5 when x = 11.

Evgen [1.6K]

Answer:

Step-by-step explanation:

12∣∣−x+3∣∣−5

Let x = 11

12∣∣−11+3∣∣−5

Do the absolute value first

12∣∣-8∣∣−5

Taking the absolute value, which means making it positive

12* 8 -5

Multiply and divide

96 -5

7 0

2 years ago