Ask question

Ask question

All categories

3 years ago

10

Two parcels of land priced at $1,200 per acre were purchased. one parcel was 15 acres in size and the other was 1 mile square in

size. how much did these two parcels cost together?

1 answer:

denis-greek [22]3 years ago

4 0

1 mile=640 acres

640 + 15=655

1,200 x 655=786,000

The answer is 786,000!

You might be interested in

Complete the statement If x/7 = 5/3, then x+7/7 =?

lawyer [7]

Lets get started :)

$\frac{x}{7} = \frac{5}{3}$
We can cross multiply, to find the value of x
3x = 7 x 5
3x = 35
We have to divide by 3 on either sides to isolate x
$\frac{3x}{3} = \frac{35}{3}$
3 and 3 cancels out

x = $\frac{35}{3}$

Then, x + $\frac{7}{7}$ ( 7 divided by 7 is 1)
$\frac{35}{3}$ + 1 = $\frac{38}{3}$

6 0

3 years ago

Your bedroom ceiling is 10 feet high and is 2/3 as high as the living room ceiling.

zaharov [31]

Let me do the work right now, then I’ll let you know!!

6 0

3 years ago

Read 2 more answers

A sculptor is selling her work at an exhibit and hopes to earn at least $31,000 in revenue. Small pieces go for $780 and large p

Sauron [17]

The answer should be

780x + 1500y=31000

please correct me if i'm wrong

3 0

3 years ago

Read 2 more answers

Find the Measure of the missing angles.

Savatey [412]

Step-by-step explanation:

1= 118°+angle 1=180°

angle 1= 180°-118°=62°

angle 2=180°-135°=45°

angle 3=135°

8 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago