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3 years ago

How do i write 658.129 in expanded form?

Mathematics

2 answers:

Orlov [11]3 years ago

5 0

Six hundred fifty eight and one hundred twenty nine thousandths

Vlada [557]3 years ago

3 0

Six hundred fifty eight and one hundred twenty nine thousandths

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I need all the help I can

Sholpan [36]

^^ they’re right so hope it helps

7 0

3 years ago

What is the surface area of a rectangular prism that has a height of 5 cm, a width of 10 cm, and a depth of 4 cm?

Ket [755]

Option C: $220 \ cm^2$ is the surface area of the rectangular prism

Explanation:

The height of the rectangular prism is 5 cm

The width of the rectangular prism is 10 cm

The depth of the rectangular prism is 4 cm

We need to determine the surface area of the rectangular prism.

The surface area of the rectangular prism can be determined using the formula,

$A=2(w l+h l+h w)$

Substituting $w=10$ , $h=5$ and $l=4$ in the formula, we get,

$A=2 \cdot(10 \cdot 4+5 \cdot 4+5 \cdot 10)$

Multiplying the terms within the bracket, we have,

$A=2 \cdot(40 +20+5 0)$

Adding all the values within the bracket, we get,

$A=2 \cdot(110)$

Multiplying, we have,

$A=220$

Thus, the surface area of the rectangular prism is $220 \ cm^2$

Hence, Option C is the correct answer.

4 0

3 years ago

Read 2 more answers

A garden table and a bench cost $781 combined. The garden table costs $69 less than the bench. What is the cost of the bench?

Arlecino [84]

Answer:

$425

Step-by-step explanation:

x = bench

x - 69 = garden table

x + (x - 69)=781

x + x - 69 =781

2x - 69 = 781

2x -69 +69 = 781 + 69

2x = 850

2x/2 = 850/2

x = 425

Bench = 425

6 0

3 years ago

Read 2 more answers

How many ways can a person with five nickels, three dimes, and two quarters make a 75-cent purchase from the machine?

DIA [1.3K]

2 quarters, 2 dimes, and 1 nickel

8 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

3 years ago