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Kazeer [188]
4 years ago
15

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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Answer asap --------------------
blsea [12.9K]

OPTION A is the correct answer

5 0
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. :)
7 0
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.
5 0
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 <em>c)</em>

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

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

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 <em>A</em> and <em>B</em> 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)

7 0
4 years ago
the hawks still have a narrow edge over the pioneers in the basketball game. they're leading 51 to 49. the pioneers have scored
scoray [572]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

<span>For the Hawks,
29 2-point shots and 0 free throws - 58 points
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25 2-point shots and 4 free throws - 54 points
</span><span>22 2-point shots and 7 free throws - 51 points (answer for Hawks)
For the Pioneers,
30 2 point shots and 0 free throws - 60 points
25 2 point shots and 5 free throws - 55 points
20 2 point shots and 10 free throws - 50 points
19 2 point shots and 11 free throws - 49 points (answer for Pioneers)</span>
5 0
3 years ago
Read 2 more answers
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