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MrRissso [65]
3 years ago
9

Y=-3+1/2x graph the system of equations and write the solution

Mathematics
1 answer:
Rudik [331]3 years ago
5 0

Answer:

(x,y) = (2.-2)

Step-by-step explanation:

Given

y = -3 + \frac{1}{2}x

-2x + y = -6

Solving (a): Graph the equations

See attachment for graph

Solving (b): The solution

This is the point of intersection between both lines

From the attached graph, the point of intersection is:

(x,y) = (2.-2)

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Sinx + siny=a<br> cosx + cosy=b<br> Find cos(x+y/2)
romanna [79]

Using the addition rule of the Sine function and the Cosine function, we obtain \cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}.

<h3>What are the formulas for (sin x + sin y) and (cos x + cos y)?</h3>
  • The formula for the addition of two Sine functions (\sin x+\sin y) is \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}.
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Given that

\sin x + \sin y = a\\\cos x + \cos y = b

Then using the above formulas, we get:

2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=a       (1)

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Dividing the equation (1) by (2), we get:

\dfrac{\sin\dfrac{x+y}{2}}{\cos\frac{x-y}{2}}=\dfrac{a}{b}\\\Longrightarrow \tan\dfrac{x+y}{2}=\dfrac{a}{b}             (3)

Now, we know that  \cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}.

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\cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\tan^2\dfrac{x+y}{2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\dfrac{a^2}{b^2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}

Therefore, using the addition rule of the Sine function and the Cosine function, we obtain \cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}.

To know more about Sine and Cosine functions, refer: brainly.com/question/27728553

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