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2 years ago

You spend $40 on 5 Pounds of concreate what is the 1 unit rate in dollars

Mathematics

2 answers:

Umnica [9.8K]2 years ago

6 0

Answer:

Step-by-step explanation:

kotegsom [21]2 years ago

3 0

Answer:

The unit rate is $8

Step-by-step explanation:

Divide 1d10t

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Eden has a part time job. She gets paid £7. 20 per hourthis week she worked for 18 1/2 hourswork out Eden's total pay for this w

Dafna1 [17]

Answer:

$133.20

Step-by-step explanation:

Since the amount of hours she worked isn't a whole number, we can first separate 18 and 1/2 and work the problem out one at a time.

1. Multiply 7.20 by 18 to get 129.60

2. Multiply 0.50 by 7.20 to get 3.60.

3. Add 129.60 and 3.60 together to get the answer, 133.20.

5 0

1 year ago

Todd’s average score for six tests was 92. If the sum of the scores of two of her tests was 188, then what was her average score

natita [175]

Answer:

Step-by-step explanation:

Todd’s average score for six tests = 92.

The sum of two of her test = 188

First, we need to find the total score for the six test. This given below:

Average = sum of all test / number of test

sum of all the test = average x number of test

average score for six tests = 92.

Number of test = 6

Sum of all the Tests = 92 x 6 = 552

Sum of four test = sum of all the test — sum of two test

Sum of four test = 552 — 188 = 364

Now we can solve for the average of the other four test as shown below:

Average of four test = 364/4= 91

3 0

2 years ago

Read 2 more answers

Is this a function or not a function? (Picture)

musickatia [10]

It is not a function

4 0

3 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

Pls answer picture is below

ahrayia [7]

Mass, hardness, density, luster

3 0

2 years ago