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3 years ago

If 35% of a natural area is to be developed, leaving 500 acres untouched, how many acres are to be developed?

2 answers:

8 0

Given that the 35% of area has been developed, the percentage of land that has not been developed is:
100-35
=65%
hence the acres of land that are to be developed will be:
35/65*500
=269.23 acres
the answer is 269.23

harkovskaia [24]3 years ago

5 0

Answer:

Area to be developed = 269.23 acres

Step-by-step explanation:

Let there is A acres natural area to be developed.

If 35% of A acres is developed then untouched area will be (100 - 35 = 65%)

Now it has been given in the question that 65% of A (untouched area) = 500 acres

$A.\frac{65}{100}=500$

$A=\frac{500}{0.65}=769.23$

So area of the natural field = 769.23 acres

Now 35% of the natural area which is to be developed = A×35%

$=769.23.\frac{35}{100}=(769.23)(.35)=269.23$

Area of the natural area to be developed is 269.23 acres

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-0.5(d - 1.5) = 2.5 What is d

yaroslaw [1]

Answer:A)0.5 We can see in the graph , that it is bell-shaped along x =2. A bell-shaped graph along one value is called symmetric graph and it represents a normal distribution.

Since, the give graph is symmetric around x=2, so the mean of the data is 2.

The point immediate left to the mean represents x-σ

so,

2 - σ = 1.5

So,

σ = 0.5

The sigma represents standard deviation.

Hence, Option A is correct ..

Step-by-step explanation:and what is on d is 2.5

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2 years ago

X - 2y < 2 graph the inequality

Korvikt [17]

Answer:

y > 1/2x - 1

First, draw the dashed line y = 1/2x - 1 (slope intercept ; y = mx + b).

Start at -1 on the y-axis, and continue going 2 units to the right, and 1 unit up for the right side of the graph.

Then starting at -1 on the y-axis, continue going 2 units to the left, and 1 unit down for the left side of the graph.

Explanation:

Convert standard form (Ax + By = C) by isolating y from the rest of the equation.

Ax + By = C → y = -Ax/B + C/B → y = mx + b.

Given a standard form equation in inequality form,

x - 2y < 2.

Set it to slope-intercept as an inequality to find the slope and y-intercept.

When negating (making opposite) a variable, you flip the inequality.

x - 2y < 2 → x - 2y - x < 2 - x → -2y < -x + 2 → 2y > x - 2 → y > 1/2x - 1.

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2 years ago

PLSSS HELP IMMEDIATELY!!!!! i’ll give brainiest, i’m not giving brainiest if u leave a link tho. (pls check whole picture!!) a

emmainna [20.7K]

Answer:

(0, 1)

Step-by-step explanation:

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2 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.

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2 years ago

Can someone explain this please???

xeze [42]

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2 years ago