Place parentheses in each equation order of operations

You use 33 cans of soup to serve 55 people. If you want to serve 2020 people, what should you do?

sveticcg [70]

First off, we need to find how much would a people need,by dividing the amount needed.

The equation of how much a people would need:

55÷33
=1.67 or 55/33 (it would be best to keep it in fraction to get a more precise answer)

To find out how much 2020 people need ,we would mutiply the answer above by 2020:

55/33×2020
≈3366.7
≈3367

Thus,you would have to prepare 3367 can of soup.

hope it helps!

3 0

3 years ago

For every 12 slices of pizza sold at Ping's Pizza Shop,

kaheart [24]

Answer:

3:5 Is the ratio because there are five pieces of cheese.

Step-by-step explanation:

If there are 12 pieces, then you can add 4 and 3 together and get 7 and then you can take 7 from 12 and get 5 which is how many cheese pieces there are. Now you know that there are 3 slices of pepperoni and 5 pieces of cheese. Therefore the ratio is 3:5.

6 0

3 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

3 years ago

Find the length of side x in simplest radical form with a rational denominator.

slega [8]

I just ask this question on the SnapCalc app the answer is in the image

4 0

3 years ago

if 3 1/2 cups of flour are used with 1 1/4 cups of brown sugar, how much sugar is used when 10 1/2 cups of flour is used?

pickupchik [31]

Step by step equation

3 0

3 years ago