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Tutor profile: Chielo A.

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Chielo A.
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Subject: Trigonometry

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Question:

Find $$\sin(\alpha+\beta)$$ and $$\cos(\alpha+\beta)$$, given that: $$\tan\alpha=-\frac{15}{8}$$, $$\sin\beta=-\frac{7}{25}$$, and neither $$P(\alpha)$$ nor $$P(\beta)$$ is in the fourth quadrant.

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Chielo A.
Answer:

The addition formulas for sine and cosine are given by: $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$ We only know the value of $$\sin\beta$$, hence we need to find $$\sin\alpha$$, $$\cos\alpha$$, and $$\cos\beta$$ from the given $$\tan\alpha=-\frac{15}{8}$$, $$\sin\beta=-\frac{7}{25}$$, and neither $$P(\alpha)$$ nor $$P(\beta)$$ is in the fourth quadrant. Recall that: $$\sin\theta=\frac{y}{r}, \cos\theta=\frac{x}{r} $$ and $$\tan\theta=\frac{y}{x} $$, where $$ r=\sqrt{x^2+y^2} $$ by pythagorean theorem. Since, $$\tan\alpha < 0$$, $$P(\alpha)$$ must be in the second quadrant (2nd Q) where $$x$$ is negative and $$y$$ is positive. Therefore, $$\tan\alpha=\frac{y}{x}=\frac{15}{-8}$$ $$\sin\alpha=\frac{+15}{\sqrt{(-8)^2+15^2}}=\frac{15}{17}$$ $$\cos\alpha=\frac{-8}{\sqrt{(-8)^2+15^2}}=-\frac{8}{17}$$ and since, $$\sin\beta < 0$$, $$P(\beta)$$ must be in the 3rd Q where both $$x$$ and $$y$$ are negative. Therefore, $$\sin\beta=\frac{y}{r}=\frac{-7}{25}$$ $$\cos\beta=\frac{-\sqrt{25^2-(-7)^2}}{25}=-\frac{24}{25}$$ Thus, $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ $$\sin(\alpha+\beta)=(\frac{15}{17})(-\frac{24}{25})+(-\frac{8}{17})(-\frac{7}{25})$$ $$\sin(\alpha+\beta)=\frac{-360}{425}+\frac{56}{425}$$ $$\sin(\alpha+\beta)=\frac{-360+56}{425}$$ $$\sin(\alpha+\beta)=-\frac{304}{425}$$ and $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$ $$\cos(\alpha+\beta)=(-\frac{8}{17})(-\frac{24}{25})-(\frac{15}{17})(-\frac{7}{25})$$ $$\cos(\alpha+\beta)=\frac{192+105}{425}$$ $$\cos(\alpha+\beta)=\frac{297}{425}$$ Therefore, $$\sin(\alpha+\beta)=-\frac{304}{425}$$ and $$\cos(\alpha+\beta)=\frac{297}{425}$$.

Subject: Chemistry

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Question:

Calculate the pressure exerted by $$ 3.07 $$ moles of a colorless unreactive gas in a steel vessel of volume $$ 4.89 $$ $$\mathsf{L}$$ at $$47 ^{\circ}\mathsf{C} $$.

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Chielo A.
Answer:

We are given the amount of gas, volume, and temperature. If we assume that the gas is an ideal gas, we can use the ideal gas equation: $$ PV = nRT $$, to solve for the pressure exerted by the gas. To solve for pressure, first, re-arrange the ideal gas equation: $$ P = \frac{nRT}{V} $$ and substitute the given values: $$n=3.07$$ moles, $$V=4.89 $$ $$\mathsf{L}$$, and $$T=47 ^{\circ}\mathsf{C} $$. We can use gas constant, $$R=0.0821 \frac{L\mathord{\cdot} atm}{K\mathord{\cdot}mol}$$, but since the given temperature is in $$^{\circ}\mathsf{C}$$, we should add $$273.15$$ to $$T$$ to convert to Kelvin, hence $$ P = \frac{(3.07 mol)(0.0821 L\mathord{\cdot} atm/K\mathord{\cdot}mol)(47+273.15)K}{4.89L} $$ $$ P = 16.50 $$ atm Therefore, the pressure exerted by the gas is 16.50 atm.

Subject: Algebra

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Question:

Find the roots of the quadratic equation $$ ax^{2} + bx + c = 0 $$ by completing the square.

Inactive
Chielo A.
Answer:

Completing the square technique can be used to solve quadratic equations that does not factor readily. To complete the square, we add a constant that will give us an equation that will yield a perfect square. To solve the given equation, we first divide the given equation by $$ a $$, to obtain $$ x^{2} + \frac{b}{a}x + \frac{c}{a} = 0 $$ then move the constant to the right side, giving $$ x^{2} + \frac{b}{a}x = - \frac{c}{a} $$ We can now complete the square by adding $$ (\frac{b}{2a})^{2} $$ to the right-hand side (RHS) and left-hand side (LHS) of the equation: $$ x^{2} + \frac{b}{a}x + (\frac{b}{2a})^{2} = - \frac{c}{a} + (\frac{b}{2a})^{2}$$ Factor LHS and simplify RHS: $$ (x+\frac{b}{2a})^{2} = \frac{-4ac + b^2}{4a^2}$$ Take the square root of both sides and solve for x: $$ x+\frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}$$ $$ x= \pm \frac{\sqrt{b^2-4ac}}{2a} -\frac{b}{2a} $$ Thus, the roots of the given equation are $$ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ This solution lead us to the quadratic formula which we can use to solve any quadratic equations of the form $$ ax^{2} + bx + c = 0 $$.

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