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4 years ago

Which one is it A. B. C. or D.

Mathematics

1 answer:

nadya68 [22]4 years ago

8 0

7x + 2y = 2
2x - y = 10

2y = - 7x + 2
- y = - 2x + 10

2y = -7x + 2
y = 2x - 10

$y=\frac{-7x+2}{2}$
y= 2x - 10

y = -3.5x + 1
y = 2x - 10

Now, let's find "x" first by doing a substituion !

2x - 10 = - 3.5x + 1
y = 2x - 10

2x + 3.5x = 1 + 10
y = 2x - 10

5.5x = 11
y= 2x - 10

$x = \frac{11}{5.5}=2$
y= 2*2 - 10 = 4 - 10 = -6

In short, the answer would be : y = -6 (Option B).

Hope this helps !

Photon

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blsea [12.9K]

OPTION A is the correct answer

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3 years ago

Can someone explain what the answer is to this and how? Thank u!!!

Sloan [31]

The answer is D.
You cannot factorise it as there are two different terms being squared with the middle term containing both. Also, if you expand any of the brackets in the questions you will not get the original equation so non of them are right so it has to be D. :)

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3 years ago

Solve for y.

storchak [24]

Subtract 1.05 from both sides. Then, you will end up with y= 6.21. So, C.

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4 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

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4 years ago

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scoray [572]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

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For the Pioneers,
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3 years ago

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