Find the value of x. A. (2x-17)" С

Mathematics

1 answer:

lora16 [44]2 years ago

8 0

The answer would be X=34

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15/2g-7=3/5 what is g

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$\frac{15}{2} g - 7 = \frac{3}{5} \\ \\ \\ 1. \: \frac{15g}{2} - 7 = \frac{3}{5} \\ 2. \: \frac{15g}{2} = \frac{3}{5} + 7 \\ 3. \: \frac{15g}{2} = \frac{38}{5} \\ 4. \: 15g = \frac{38}{5} \times 2 \\ 5. \: 15g = \frac{76}{5} \\ 6. \: g = \frac{ \frac{76}{5} }{15} \\ 7. \: g = \frac{76}{5 \times 15} \\ 8. \: g = \frac{76}{75}$

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$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\\\
cos(2\theta)=
\begin{cases}
\boxed{cos^2(\theta)-sin^2(\theta)}\\
1-2sin^2(\theta)\\
2cos^2(\theta)-1
\end{cases}\\\\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-------------------------------\\\\$

$\bf 6y+y^2=x^2\implies \cfrac{6y}{x^2}+\cfrac{y^2}{x^2}=1\implies \cfrac{6rsin(\theta )}{[rcos(\theta )]^2}+\cfrac{[rsin(\theta )]^2}{[rcos(\theta )]^2}=1
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\cfrac{6rsin(\theta )}{r^2cos^2(\theta )}+\cfrac{r^2sin^2(\theta )}{r^2cos^2(\theta )}=1\implies 
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$\bf \cfrac{6sin(\theta )}{rcos^2(\theta )}=1-\cfrac{sin^2(\theta )}{cos^2(\theta )}\implies \cfrac{6sin(\theta )}{1-\frac{sin^2(\theta )}{cos^2(\theta )}}=rcos^2(\theta )
\\\\\\
\cfrac{6sin(\theta )}{cos^2(\theta )\left[ 1-\frac{sin^2(\theta )}{cos^2(\theta )} \right]}=r\implies \cfrac{6sin(\theta )}{cos^2(\theta )-sin^2(\theta )}=r
\\\\\\
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