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tester [92]
3 years ago
14

For the following inequality, indicate whether the boundary line should be dashed or solid. 3x ≥ y + 1

Mathematics
1 answer:
Kay [80]3 years ago
8 0
The boundary line should be solid because it have a 'great/less than or equal to' sign
You might be interested in
Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi
kompoz [17]

Answer:

The general solution is

y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

     + \frac{x^2}{2}

Step-by-step explanation:

Step :1:-

Given differential equation  y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

(D^4 -2D^3+D^2)y = e^x+1

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

                         m^4 -2m^3+m^2 = 0

                         m^2(m^2 -2m+1) = 0

(m^2 -2m+1) this is the expansion of (a-b)^2

                        m^2 =0 and (m-1)^2 =0

The roots are m=0,0 and m =1,1

complementary function is y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x

<u>Step 2</u>:-

The particular equation is    \frac{1}{f(D)} Q

P.I = \frac{1}{D^2(D-1)^2} e^x+1

P.I = \frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}

P.I = I_{1} +I_{2}

\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}

\frac{1}{D} means integration

\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx

applying in integration u v formula

\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx  } )\, dx

I_{1} = \frac{1}{D^2(D-1)^2} e^x

\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))

\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

I_{2}= \frac{1}{D^2(D-1)^2}e^{0x}

\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x

again integration  \frac{1}{D} x = \frac{x^2}{2!}

The general solution is y = y_{C} +y_{P}

         y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

      + \frac{x^2}{2!}

3 0
3 years ago
How do i make this fraction (7/3) a mixed number?
uranmaximum [27]

You get the whole numbers (1's) out of it and show the left over fraction next to it. (e.g. there is 7/3 as a mixed number is 2 1/3 )

7 0
3 years ago
Will mark you as brainliest. If your good with polynomials.
kobusy [5.1K]
To solve/simplify this all you have to do is group like terms (the x^2's with each other, the x's with each other, and the normal numbers, -8)

14x^2-8+5x-6x^2+2x

group the x^2 (add 14x^2 to -6x^2)

8x^2-8+5x+2x

group the x's together (add 5x and 2x together)

8x^2+7x-8

Your answer will be d) 8x^2+7x-8



4 0
3 years ago
Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm
Vlad1618 [11]

Answer:

Confidence =1-\alpha=1-0.01=0.99

And then the confidence level would be given by 99%

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

X \sim N(\mu, \sigma)

The distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar x =512 represent the sample mean

\sigma=100 represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}   (1)

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}

And we can find the critical value z_{\alpha/2} like this:

z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576

And we know that on the right tail of the z score =2.576 we have \alpha/2 of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998=\alpha/2, so then \alpha=0.00498*2=0.009995 \approx 0.01, and then we can find the confidence like this:

Confidence =1-\alpha=1-0.01=0.99

And then the confidence level would be given by 99%

8 0
3 years ago
What is the value of b in the equation 3b + 2(b − 1) = 8?
Ray Of Light [21]
<span>3b + 2(b − 1) = 8
3b + 2b - 2 = 8
5b = 10
b = 2

answer b = 2</span>
7 0
3 years ago
Read 2 more answers
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