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atroni [7]
3 years ago
5

5.4.4 Practice Modeling: Two variable systems of inequalities

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

<h2>The system has no solution, because there's no intersected areas.</h2>

Step-by-step explanation:

We have the following inequalities

2x-3y>6

y \geq x^{2} -1

This is a two variable system of inequalities, where the first one is linear (line) and the second one is quadratic (parabolla).

It's important to know that a two variable inequalitiy has ordered pairs as solution, which means its solution is an area in the coordinate system. To find such area, we just need to graph both expressions as equations:

2x-3y=6\\y=x^{2} -1

(First image attached)

Then, we use the inequality signs to find each area of solution, as the second image shows.

In this case, the system has no solution, because there's no intersected areas.

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Answer:

3/11

Step-by-step explanation:

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What's X? to: <br> 15x - 7x = (-52) - 5x
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-4

Because 15x-7x= 8x
8x+5x= 13x
13x= -52
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A right triangle is shown below. What is cos EFD?<br> Will make BRAINLIEST
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There is $500 in Holly's bank account. She takes out $50 from her account each month but doesn't put money back in. What is the
galina1969 [7]

Answer:

The rate of change or slope m  = -50

Step-by-step explanation:

The slope-intercept form of the line equation

y = mx+b

where

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  • b is the y-intercept

Given that there is $500 in Holly's bank account.

Thus, the y-intercept b = 500

It is stated that she takes out $50 from her account each month but doesn't put money back in.

It means the rate of change or slope is decreasing.

Thus, the rate of change or slope m  = -50

<u>EXTRA BONUS!</u>

We can also determine the line equation by substituting m = -50 and b = 500 in the slope-intercept form

y = mx+b

y = -50x + 500

Here:

The slope m = -50

The y-intercept = 500

3 0
3 years ago
Find the Fourier series of f on the given interval. f(x) = 1, ?7 &lt; x &lt; 0 1 + x, 0 ? x &lt; 7
Zolol [24]
f(x)=\begin{cases}1&\text{for }-7

The Fourier series expansion of f(x) is given by

\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7

where we have

a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx
a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)
a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2

The coefficients of the cosine series are

a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx
a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)
a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}
a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}

When n is even, the numerator vanishes, so we consider odd n, i.e. n=2k-1 for k\in\mathbb N, leaving us with

a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}

Meanwhile, the coefficients of the sine series are given by

b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx
b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)
b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}
b_n=\dfrac{7(-1)^{n+1}}{n\pi}

So the Fourier series expansion for f(x) is

f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7
3 0
3 years ago
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