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3 years ago

Which of the following is a solution of the system x – 2y < 4 and y > – 2x – °5?

Mathematics

1 answer:

Nutka1998 [239]3 years ago

8 0

Answer: B

Step-by-step explanation:

Given the two systems of inequality to be:

x – 2y < 4

Rearrange the equation,

X - 2y < 4

-2y < 4 - x

The sign will change as we divide both sides by -2

Y > -2 + x/2

Y > x/2 - 2 ...... (1)

and

y > – 2x – 5 ..... (2)

Multiply (1) by 2 and (2) by 1/2

2y > x - 4

y/2 > -x - 5/2

Since the inequality sign of the two are the same, you can eliminate x by addition.

2 1/2 y > - 4 5/2

Change mixed fraction to improper fraction.

5y/2 > -13/2

Cross multiply

10y > -26

y > -26/10

y > - 2.6

Base on the value got for y, the correct answer is B ( -8, 2 ) because only option B and D has negative value for x and y is not equal to zero.

You confirm by plugging in the value in the two inequality equations.

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A and B together complete a work in 20days. They worked together for 5days and B leaves. How long will A takes to finish the rem

andreev551 [17]

Answer:

27 days

Step-by-step explanation:

A alone, rate per day = 1/36

A and B together, rate per day = 1 / 20

Total Workdone for 5 days :

1/ 20 * 5 = 5 / 20 = 1/4

Fraction of work left :

1 - 1/4 = 3/4

3 / 4 * 36 /1

= (3 * 36) / 4

= 108 / 4

= 27 days

3 0

3 years ago

What is 1/3 minus 1/4

Fed [463]

Answer:

1/3 - 1/4 = 1/12.

Step-by-step explanation:

5 0

3 years ago

Move each inequality next to the number line it represents.<br> Number Line<br> Inequality

Art [367]

1. x > -2
2. x ≤ -2
3. x ≥ -2
4. x < -2

5 0

3 years ago

Place these numbers on the number line. 1/4, 2/3, 3/4, 6/8

Lelu [443]

The best way to put them into a number line is to make sure that they've all got the same denominator, which in this case, will be 24, as this is the lowest common multiple of 3,4 and 8. You will then need to multiply these fractions accordingly. The first one (1/4) will need to be multiplied by 6 to get a denominator of 24, so this will then give you a result of 6/24. The second one (2/3), will need to be multiplied by 8, which will then give you a result of 16/24. The third one (3/4)will also need to be multiplied by 6, giving you 18/24. The final one (6/8) will need to be multiplied by 3, giving you 18/24. You will then need to put these in order, which will be- 1/4, 2/3, 3/4, and 6/8. The alternative method is to turn these into decimals- 1/4 is equivalent to 25%, which is equal to 0.25. 2/3 is equivalent to 66.66%, which is equivalent to 0.66. 3/4 is equivalent to 75%, which is equal to 0.75. 6/8 is also equivalent to 75%, which is equal to 0.75, you can then easily line them up this way. Hope this has been able to help you

4 0

4 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

4 years ago