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daser333 [38]
3 years ago
15

Please help me pleeeeeeaaassseee

Mathematics
1 answer:
mr_godi [17]3 years ago
4 0

Answer:

Step-by-step explanation:

answer no 3:

1] Factor out the common term 2.

2(3x+y) = -26

2] Divide both sides by 2.

3x+y= -26/2

3x+y = − 13

 

3]Subtract y from both sides.

3x = - 13-y

4] Divide both sides by 3.

x = -13-y/3

Part 2:-

Add 6y to both sides.

x=21+6y

x=21+6y

​  

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pashok25 [27]

Option A:

\tan(105^\circ)=-(2+\sqrt{3})

Solution:

<u>To evaluate tan(105)°:</u>

105° can be written as sum of 60° and 45°.

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Using the summation identity:

$\tan (x+y)=\frac{\tan (x)+\tan (y)}{1-\tan (x) \tan (y)}

$\tan \left(105^{\circ}\right)=\frac{\tan \left(45^{\circ}\right)+\tan \left(60^{\circ}\right)}{1-\tan \left(45^{\circ}\right) \tan \left(60^{\circ}\right)}

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Substitute this in the above equation.

              $=\frac{1+\sqrt{3}}{1-1 \cdot \sqrt{3}}

              $=\frac{1+\sqrt{3}}{1-\sqrt{3}}

To rationalize the denominator multiply by the conjugate \frac{1+\sqrt{3}}{1+\sqrt{3}}.

              $=\frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}

Using exponent formula: a^{b} \cdot a^{c}=a^{b+c} and (x-y)(x+y)=x^2-y^2

              $=\frac{(1+\sqrt{3})^2}{(1^2-(\sqrt{3})^2)}

Using exponent formula: (a+b)^{2}=a^{2}+2 a b+b^{2}

              $=\frac{1^{2}+2 \cdot 1 \cdot \sqrt{3}+(\sqrt{3})^{2}}{1-3}

              $=\frac{4+2 \sqrt{3}}{-2}

              $=\frac{2(2+ \sqrt{3})}{-2}

              =-(2+\sqrt{3})

\tan(105^\circ)=-(2+\sqrt{3})

Hence option A is the correct answer.

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