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fenix001 [56]
4 years ago
12

3x 2 + 2x - 5 and -4 + 7x 2

Mathematics
2 answers:
fiasKO [112]4 years ago
4 0

Answer:

16+2x-5adn

Step-by-step explanation:

6+2x-5adn-4+14. solution: is 16=2x-5adn

vladimir1956 [14]4 years ago
4 0

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{10 {x}^{2}  + 2x - 9}}}}}

Step-by-step explanation:

\sf{3 {x}^{2}  + 2x - 5 + ( - 4 + 7 {x}^{2} )}

When there is a ( + ) in front of an expression in parentheses, there is no need to change the sign of each term.

That means , the expression remains the same.

⇒\sf{3 {x}^{2}  + 2x - 5 - 4 + 7 {x}^{2} }

Collect like terms

⇒\sf{ {10x}^{2}  + 2x - 5 - 4}

Calculate

⇒\sf{10 {x}^{2}  + 2x - 9}

Hope I helped!

Best regards!!

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The circle was 126 meters around.

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Step 1; It is given that the big circle was 40 feet across the center. This implies the distance from any particular point on the circle to any other point is 40 feet. This means the diameter of this given circle is 40 feet. Any given circle's radius is 0.5 times the diameter of that same circle. So we divide 40 by 2 to get the circle's radius. So the radius is 40/2=20 feet. So the radius of this given circle is 20 feet.

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Find the exact value of the trigonometric expression.
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\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
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\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}

\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}


which simplifies thus far to 

\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
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2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
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2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
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2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
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2\sqrt{2-\sqrt{3}}
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