[6kg-(5kg)(sin30)-(5kg)(cos30)(0.2)](9.8m/s^2) / 5kg + 6kg

asked by guest
on Sep 22, 2024 at 6:23 am



You asked:

Evaluate the expression: \(\frac{\left(6 k g - 5 k g \cdot \sin\left( 30 \right) - 5 k g \cdot \cos\left( 30 \right) \cdot 0.2\right) \cdot \frac{9.8 m}{{s}^{2}}}{5 k g} + 6 k g\)

MathBot Answer:

Evaluated



\(\displaystyle \frac{\left(6 k g - 5 k g \cdot \sin\left( 30 \right) - 5 k g \cdot \cos\left( 30 \right) \cdot 0.2\right) \cdot \frac{9.8 m}{{s}^{2}}}{5 k g} + 6 k g = 6 g k + \frac{49 m \left(- g k \cos{\left(30 \right)} - 5 g k \sin{\left(30 \right)} + 6 g k\right)}{25 s^{2} g k} \)

Expanded

\[\frac{\left(6 k g - 5 k g \cdot \sin\left( 30 \right) - 5 k g \cdot \cos\left( 30 \right) \cdot 0.2\right) \cdot \frac{9.8 m}{{s}^{2}}}{5 k g} + 6 k g = 6 g k - \frac{49 m \cos{\left(30 \right)}}{25 s^{2}} - \frac{49 m \sin{\left(30 \right)}}{5 s^{2}} + \frac{294 m}{25 s^{2}}\]

Factored

\[\frac{\left(6 k g - 5 k g \cdot \sin\left( 30 \right) - 5 k g \cdot \cos\left( 30 \right) \cdot 0.2\right) \cdot \frac{9.8 m}{{s}^{2}}}{5 k g} + 6 k g = \frac{150 s^{2} g k - 49 m \cos{\left(30 \right)} - 245 m \sin{\left(30 \right)} + 294 m}{25 s^{2}}\]

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