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Tutor profile: Mike R.

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Mike R.
After-school tutor for mathematics.
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Questions

Subject: Trigonometry

TutorMe
Question:

$$1)$$ If $$\cos(x) = \frac{60}{61}$$ and $$\cot(x) = \frac{60}{11}$$, find $$\sin(x)$$. $$2)$$ Show that $$\frac{\cot(x)}{\csc(x)} = \cos(x)$$. $$3)$$ Simplify: $$\cos(x) + \sin(x) \tan(x)$$

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Mike R.
Answer:

$$1)$$ $$\cot(x) = \frac{60}{11} \Rightarrow \frac{1}{\tan(x)} = \frac{60}{11} \Rightarrow \frac{\cos(x)}{\sin(x)} = \frac{60}{11} \Rightarrow \frac{\frac{60}{61}}{\sin(x)} = \frac{60}{11} \Rightarrow \frac{60}{61} = \frac{60}{11} \sin(x) \Rightarrow \sin(x) = \frac{60}{61} \cdot \frac{11}{60} = \frac{11}{61}$$ $$2)$$ $$\frac{\cot(x)}{\csc{x}} = \frac{\frac{\cos(x)}{\sin(x)}}{\frac{1}{\sin(x)}} = \frac{\cos(x)}{\sin(x)} \cdot \sin(x) = \cos(x)$$ $$3)$$ $$\cos(x) + \sin(x) \tan(x) = \cos(x) + \sin(x) \cdot \frac{\sin(x)}{\cos(x)} = \frac{\cos^2(x)}{\cos(x)} + \frac{\sin^2 (x)}{\cos(x)} = \frac{\cos^2(x) + \sin^2(x)}{\cos(x)} = \frac{1}{\cos(x)} = \sec(x)$$

Subject: Calculus

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Question:

$$1)$$ Compute the derivative $$y'$$ of the function $$y = x^x$$. $$2)$$ Compute the integral $$\int_0^{\pi} \sin(x) + \cos(x) dx$$ $$3)$$ Calculate the implicit derivative $$y'$$ of $$x^2 + y^2 = 2xy$$

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Mike R.
Answer:

$$1)$$ $$y = x^x \Rightarrow \ln(y) = \ln (x^x) \Rightarrow \ln(y) = x \ln(x) \Rightarrow (\ln(y))' = (x \ln(x))' \Rightarrow \frac{1}{y} y' = x \cdot \frac{1}{x} + \ln(x) \cdot 1 = 1 + \ln(x) \Rightarrow \frac{1}{y}y' = 1 + \ln(x) \Rightarrow y' = y(1 + \ln(x)) = x^x(1 + \ln(x))$$. Thus, $$y' = x^x(1 + \ln(x))$$. $$2)$$ $$\int_0^{\pi} \sin(x) + \cos(x) dx = [-\cos(x) + \sin(x)]_0^\pi = [-\cos(\pi) + \sin(\pi)] - [- \cos(0) + \sin(0)] = [-(-1)+ 0]-[-1 + 0] = 1 - (-1) = 1 + 1 = 2$$ $$3)$$ $$(x^2 + y^2)' = (2xy)' \Rightarrow 2x + 2yy' = 2xy' + 2y \Rightarrow 2yy' = 2xy' + 2y - 2x \Rightarrow 2yy' - 2xy' = 2y - 2x \Rightarrow y'(2y - 2x) = 2y - 2x \Rightarrow y' = 1$$.

Subject: Algebra

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Question:

$$1)$$ Determine the sum of $$\frac{1}{3\sqrt{25}} + \frac{1}{2 \sqrt[3]{27}}$$ $$2)$$ Evaluate the expression into simplest form: $$\frac{8x^6 z^4 + 4x^4 z^2}{4x^2 z}$$ $$3)$$ Determine the product of $$\sqrt{-6}(\sqrt{-4} - \sqrt{3})$$

Inactive
Mike R.
Answer:

$$1)$$ $$\frac{1}{3\sqrt{25}} + \frac{1}{2 \sqrt[3]{27}} = \frac{1}{3 \cdot 5} + \frac{1}{2 \cdot 3} = \frac{1}{15} + \frac{1}{6} = \frac{2}{30} + \frac{5}{30} = \frac{7}{30}$$ $$2)$$ $$\frac{8x^6 z^4 + 4x^4 z^2}{4x^2 z} = \frac{4x^4 z^2(2x^2z^2 + 1)}{4x^2 z} = x^2 z(2x^2 z^2 + 1) = 2x^4 z^3 + x^2z$$ $$3)$$ $$\sqrt{-6}(\sqrt{-4} - \sqrt{3}) = i\sqrt{6}(i \sqrt{4} - \sqrt{3}) = i \sqrt{6}(2i - \sqrt{3}) = (i \sqrt{6})(2i) + (-\sqrt{3})(i \sqrt{6}) = -2 \sqrt{6} + (-\sqrt{3})(i \sqrt{3} \cdot \sqrt{2}) = -2 \sqrt{6} + (-3i \sqrt{2}) = -2\sqrt{6} - 3i\sqrt{2}$$

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