Aggh please help with this

At the start of the month, Jodie had sold 885 copies of her new book. At the end of the month, she had sold 1,364 copies of her

ozzi

Approximately $ 31419 profit is earned from selling books in entire month

Solution:

Given that Jodie had sold 885 copies of her new book

At the end of the month, she had sold 1,364 copies of her book

We have to determine the profit earned in the entire month

From given information,

Start of month sale = 885 copies

End of month sale = 1364 copies

Total copies of books sold = Start of month sold + end of month sold

Total copies of books sold = 885 + 1364 = 2249

Also given that each book profit is $ 13.97

Profit of 1 book = $ 13.97

Therefore for 2249 books, profit earned is given as:

Profit of 2249 books = $ 13.97 x 2249 = 31418.53

Therefore approximately $ 31419 profit is earned from selling books in entire month

5 0

2 years ago

Can anybody give real answers please? Thanks.

Kazeer [188]

I believe it is 17.0, I hope this helps some

8 0

3 years ago

How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x

Fofino [41]

Answer:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

Step-by-step explanation:

We are given the function:

$f(x)=-4x^5-8x^3+12x$

And we want to finds its zeros.

Therefore:

$0=-4x^5-8x^3+12x$

Firstly, we can divide everything by -4:

$0=x^5+2x^3-3x$

Factor out an x:

$0=x(x^4+2x^2-3)$

This is in quadratic form. For simplicity, we can let:

$u=x^2$

Then by substitution:

$0=x(u^2+2u-3)$

Factor:

$0=x(u+3)(u-1)$

Substitute back:

$0=x(x^2+3)(x^2-1)$

By the Zero Product Property:

$x=0\text{ and } x^2+3=0\text{ and } x^2-1=0$

Solving for each case:

$x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}$

Therefore, our real and complex zeros are:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

4 0

2 years ago

Betty invests $45,000 into a cd which has an interest rate of 6.3% that compounds monthly for a total of 5 years. how much will

Lelechka [254]

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$45000\\ r=rate\to 6.3\%\to \frac{6.3}{100}\dotfill &0.063\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &5 \end{cases} \\\\\\ A=45000\left(1+\frac{0.063}{12}\right)^{12\cdot 5}\implies A=45000(1.00525)^{60}\implies A\approx 61611$

5 0

1 year ago

"A serving of fish contains 50 g protein and 4 g of fat. If protein has a caloric value of 4 kcal/g and fat has 9 kcal/g, how ma

evablogger [386]

Answer: 236 kcal

Step-by-step explanation:

Given : A serving of fish contains 50 g protein and 4 g of fat.

i.e. content of protein = 50 g

content of fat = 4 g

Also, The caloric value in protein = 4 kcal/g

The caloric value in fat = 9 kcal/g

Calories in 50 g protein = (content of protein)x (caloric value in protein )

= 50g x 4 kcal/g = 200 kcal

Calories in 4 g fat= (content of fat)x (caloric value in fat)

=4 g x 9 kcal/g = 36 kcal

Then, the total calories in the serving will be :Calories in 50 g protein + Calories in 4 g fat

= 200 kcal+ 36 kcal =236 kcal

Hence, there are 236 kcal in the serving.

7 0

3 years ago