A forest ranger can see for a distance of 12 miles from a firetower.How many square miles can he observe?

Last year, a women's professional organization made two small-business loans totaling $28,000 to young women beginning their own

igor_vitrenko [27]

Answer:

$7,000 at a rate of 7% and $21,000 at a rate of 14%.

Step-by-step explanation:

Let x be amount invested at 7% and y be amount invested at 14%.

We have been given that a women's professional organization made two small-business loans totaling $28,000. We can represent this information in an equation as:

$x+y=28,000...(1)$

The interest earned at 7% in one year would be $0.07x$ and interest earned at 14% in one year would be $0.14x$ .

We are also told that the organization received from these loans was $3,430. We can represent this information in an equation as:

$0.07x+0.14y=3,430...(2)$

Form equation (1), we will get:

$x=28,000-y$

Upon substituting this value in equation (2), we will get:

$0.07(28,000-y)+0.14y=3,430$

$1960-0.07y+0.14y=3,430$

$1960+0.07y=3,430$

$1960-1960+0.07y=3,430-1960$

$0.07y=1470$

$\frac{0.07y}{0.07}=\frac{1470}{0.07}$

$y=21,000$

Therefore, an amount of $21,000 was invested at a rate of 14%.

$x=28,000-y$

$x=28,000-21,000$

$x=7,000$

Therefore, an amount of $7,000 was invested at a rate of 14%.

8 0

3 years ago

How do you express to 24.357 in unit form

tamaranim1 [39]

Twenty ten thousandths, Four thousandths, Three hundredths,Five tens,Seven Ones

6 0

3 years ago

Read 2 more answers

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$