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Darina [25.2K]
3 years ago
10

1/3(-12k+30)=6+15k Solve for k

Mathematics
2 answers:
almond37 [142]3 years ago
8 0

Answer: k = 4/9

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Aneli [31]3 years ago
5 0

Answer:

k = 4/19

Step-by-step explanation:

1/3 (- 12k + 30) = 6 + 15k

- 12/3k + 30/3 = 6 + 15k

- 12/3k - 15k = 6 - 30/3

- 12/3k - 15/1k = 6/1 - 30/3

- 12/3k - 45/3k = 18/3 - 30/3

- 57/3k = - 12/3

- k = - 12/3 / 57/3

- k = - 12/3 * 3/57

- k = - 12/57

k = 12/57

k = 4/19

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Find the exact value of sin(cos^-1(4/5))
boyakko [2]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2762144

_______________


Let  \mathsf{\theta=cos^{-1}\!\left(\dfrac{4}{5}\right).}


\mathsf{0\le \theta\le\pi,}  because that is the range of the inverse cosine funcition.


Also,

\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}


Square both sides and apply the fundamental trigonometric identity:

\mathsf{(5\,cos\,\theta)^2=4^2}\\\\
\mathsf{5^2\,cos^2\,\theta=4^2}\\\\
\mathsf{25\,cos^2\,\theta=16\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{25\cdot (1-sin^2\,\theta)=16}

\mathsf{25-25\,sin^2\,\theta=16}\\\\
\mathsf{25-16=25\,sin^2\,\theta}\\\\
\mathsf{9=25\,sin^2\,\theta}\\\\
\mathsf{sin^2\,\theta=\dfrac{9}{25}}


\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{3}{5}}


But \mathsf{0\le \theta\le\pi,} which means \theta lies either in the 1st or the 2nd quadrant. So \mathsf{sin\,\theta} is a positive number:

\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\
\therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}


I hope this helps. =)


Tags:  <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>

3 0
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