1. What is the change per week in the amount of money in the account

PLEASE HELP ME TO ANSWER THIS QUESTION THANK YOU!

iVinArrow [24]

Answer:

Step-by-step explanation:

a. 4cm * 9cm * 2cm = 72 cm^3

b. 5m * 4m * 9m = 180 m^3

c. 10dm * 36 dm^2 (because the square could also be represented as 6cm * 6cm) =360 dm^3

d. (56mm)^3 = 175616 mm^3

e. 10 m * 1 m^2 (because of the square) = 10m^3

3 0

1 year ago

Read 2 more answers

Consider the inequality -20.2 > 0y. Which of the following must be true? Check all that apply.

Lunna [17]

Answer:

1. You cannot divide by zero.

2. The inequality is equivalent to -20.2 > 0, which is false.

5 0

3 years ago

Your drapery business has just gotten a new client who needs 4 drapes made. Each one will need to be 8 feet by 5 feet and you mu

UNO [17]

22 because 6 inches on top and bottom means you need 12 inchesw each drape and you have 4 drapes so 48 inches therefor rounding that you get 1 yard each added onto the 8 feet and 5 feet for each drape.

6 0

3 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago

A person sold 100 shares of a stock at a loss of 40%. If the selling price for the 100 shares was $3,000, which of the following

Kisachek [45]

A person sold 100 shares of a stock at a loss of 40%.

selling price for the 100 shares was $3,000

Let the stock was bought at $x then we can write

$x-\frac{40x}{100}=3000\\ \\ \frac{100x-40x}{100}=3000\\ \\ \ \frac{60x}{100}=3000\\ \\ \ 60x= 3000*100\\ \\ x=\frac{3000*100}{60} \\ \\ \ x=5*100=5000\\$

Hence Amount paid for the Stock was $5000

7 0

3 years ago