HELP ILL GIVE BRAINLISTTT!

After a 75% reduction, you purchase a new clothes dryer on sale for $125. What was the original price of the clothes dryer?

Free_Kalibri [48]

The clothes dryer would've been $500 originally :)

7 0

3 years ago

What is the value of x in the equation 3(2x + 8) = 0? Group of answer choices −8 −4 4 8 please hurry!

Dafna11 [192]

Answer:

-4

Step-by-step explanation:

3(2x+8) = 0

Distribute

6x + 24 = 0

24 = -6x + 0

24/-6 = -4

8 0

3 years ago

A certificate of deposit has an annual simple interest rate of 5.25%. If $567 in interest is earned over a 6 year period, how mu

Scrat [10]

Find the the amount of interest per year. Since $567 were earned over 6 years, you divide the interest earned by the number of years it took to accumulate it:

$567/6 years = $94.5/year

Divide that by the interest rate which the interest rate needs to be in decimal form:

$94.5/.0525 = $1800

4 0

4 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

If k is a positive integer and n = k(k + 7), is n divisible by 6 ? (1) k is odd. (2) When k is divided by 3, the remainder is 2.

olganol [36]

Answer:

1.) Yes

2.) Yes

Step-by-step explanation:

Given that

n = k(k + 7)

If k is a positive integer and n = k(k + 7), is n divisible by 6 ?

(1) k is odd. Yes.

Let assume that k = 3

Then, n = 3(3 + 7)

n = 3 × 10

n = 30.

30 is divisible by 6.

(2) When k is divided by 3, the remainder is 2. That is,

Let k = 5

Then,

5/3 = 1 remainder 2

Substitute k into the equation

n = k(k + 7)

n = 5(5 + 7)

n = 5 × 12

n = 60

And 60 is divisible by 6.

Therefore, the answer to both questions is Yes.

5 0

3 years ago