All categories

2 years ago

The size of a typical city park is about 5 acres, or 217,800 square feet. If a developer plans to

Mathematics

1 answer:

pshichka [43]2 years ago

3 0

Using the percentage concept, it is found that 95.4% of the park will be used for other activities.

<h3>What is a percentage?</h3>

The percentage of an amount a over a total amount b is given by a multiplied by 100% and divided by b, that is:

$P = \frac{a}{b} \times 100\%$

In this problem, there will be two courts of 5,000 square feet each, totaling 10,000 square feet. The remainder 207,800 square feet will be used for other activities, hence the percentage is given by:

$P = \frac{207,800}{217,800} \times 100\% = 95.4\%$

95.4% of the park will be used for other activities.

More can be learned about the percentage concept at brainly.com/question/10491646

You might be interested in

A chef makes 1 and one half gallons of soup in a large pot.how many 1-cup servings can the chef get from this large pot of soup

agasfer [191]

16cups because 1 gallon + 16 cups

8 0

3 years ago

brainly is the best app in the whole world) sam has 5 dollars he works for 5 hours and he earns 900 dollars and hour how much mo

saveliy_v [14]

Answer:

Sam has $900 more money. In total he now has $905.

5 0

2 years ago

Volume of a right rectangular prisms the side lengths are 1ft,2ft,3ft,4ft,and 5ft.

In-s [12.5K]

The volume would be 120sqft.

6 0

3 years ago

Read 2 more answers

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{

il63 [147K]

Answer:

$\frac{x-3}{2x-3}$ . hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

$x^2-5x+6$

Now let's find what ac is (it's just c since a=1...)

$AC= 6$

List factors of -6

$\pm1, \pm2, \pm3, \pm6$ .

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

$(x-2)(x-3)$

Ok now let's factor the second equation:

$2x^2-7x+6$

Multiply a and c

$AC = 12$

List factors of 12:

$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ .

Factors that add up to -7 and multiply to 12:

$-3\ and\ -4$

Rewrite equation:

$2x^2-4x-3x+6$

Group terms:

$(2x^2-4x)+(-3x+6)$

Factor out GCF:

$2x(x-2)-3(x-2)$

Rewrite:

$(2x-3)(x-2)$

Now let's write out the equation using these factors:

$\frac{(x-2)(x-3)}{(2x-3)(x-2)}$ .

Here we can factor out the x-2 and the simplified form is:

$\frac{x-3}{2x-3}$

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0

2 years ago